Either statement alone is sufficient.

From either statement alone, it can be determined that \(\displaystyle m \angle A = 30^{\circ }\) and \(\displaystyle m \angle C = 60^{\circ }\); each statement gives one angle measure, and the other can be calculated by subtracting the first from \(\displaystyle 90^{\circ }\), since the acute angles of a right triangle are complementary. 

Also, since \(\displaystyle \angle B\) is the right angle, \(\displaystyle \overline{AC}\) is the hypotenuse, and \(\displaystyle \overline{BC}\), opposite the \(\displaystyle 30 ^{\circ }\) angle, the shorter leg of a 30-60-90 triangle. From either statement alone, the 30-60-90 Theorem can be used to find the length of longer leg \(\displaystyle \overline{AB}\). From Statement 1 alone, \(\displaystyle \overline{AB}\) has length \(\displaystyle \frac{\sqrt{3}}{2}\) times that of the hypotenuse, or \(\displaystyle \frac{\sqrt{3}}{2} \cdot 22 = 11\sqrt{3}\). From Statement 2 alone, \(\displaystyle \overline{AB}\) has length \(\displaystyle \sqrt{3}\) of the shorter leg, or \(\displaystyle 11\sqrt{3}\).

This is a Pythagorean theorem question.  The lengths of a right triangle are related by the following equation:  \(\displaystyle a^{2}+b^{2}=c^{2}.\)  In the problem statement, \(\displaystyle c=13\textup{in}\) and \(\displaystyle a=5\textup{in.}\) Therefore, \(\displaystyle b^{2}= c^{2}-a^{2}= 13^{2}-5^{2}=169-25=144. b=\sqrt{144}=12\textup{in.}\)

All right angles are congruent, so \(\displaystyle \small \small \small \angle B \cong \angle E\).

Since Statement 1 tells us that \(\displaystyle \small \small \angle A \cong \angle D\), this sets up the conditions for the Angle-Angle Similarity Postulate, so \(\displaystyle \small \small \Delta ABC \sim \small \Delta DEF\).

Statement 2 alone only tells us their hypotenuses. Congruence between one pair of angles and the measures of one pair of sides is insufficient information to determine whether two triangles are similar (given one angle, at least two pairs of proportional sides are required).

Therefore, the answer is that Statement 1 alone, but not Statement 2, is sufficient.

Each statement alone only gives a relationship between two sides within one triangle, so neither alone answers the question of the similarity of the two triangles.

Assume both statements are true. Then, since \(\displaystyle XZ < YZ\),  \(\displaystyle \frac{1}{ XZ}> \frac{1}{YZ}\). 

By the multiplication property of inequality, since 

\(\displaystyle AC > BC\) and \(\displaystyle \frac{1}{ XZ}> \frac{1}{YZ}\),

\(\displaystyle AC \cdot \frac{1}{ XZ} > BC \cdot \frac{1}{YZ}\)

\(\displaystyle \frac{AC}{ XZ} > \frac{BC}{YZ}\)

Since, by definition, \(\displaystyle \triangle ABC \sim \triangle XYZ\) requires that \(\displaystyle \frac{AC}{ XZ} = \frac{BC}{YZ}\), \(\displaystyle \triangle ABC \nsim \triangle XYZ\).

Assume Statement 1 alone. The ratio of the perimeters does not in and of itself establish similarity, since only one angle congruence is known.

Assume Statement 2 alone. The equation can be rewritten as a proportion statement:

\(\displaystyle AC \cdot YZ = BC \cdot XZ\)

\(\displaystyle \frac{AC \cdot YZ }{BC \cdot YZ}= \frac{BC \cdot XZ}{BC \cdot YZ }\)

\(\displaystyle \frac{AC }{BC }= \frac{ XZ}{ YZ }\)

This establishes that two pairs of corresponding sides are in proportion. Their included angles are both right angles, so \(\displaystyle \angle C \cong \angle Z\), and \(\displaystyle \triangle ABC \sim \triangle XYZ\) follows from the Side-Angle-Side Similarity Theorem.

Assume Statement 1 alone. The statement used to find a sidelength ratio:

\(\displaystyle 3 \cdot AB = 4 \cdot XY\)

\(\displaystyle \frac{3 \cdot AB}{3 \cdot XY} = \frac{4 \cdot XY}{3 \cdot XY}\)

\(\displaystyle \frac{ AB}{ XY} = \frac{4 }{3 }\)

However, since we only know one sidelength ratio, similarity cannot be proved or disproved.

From Statement 2, another ratio can be found:

\(\displaystyle 4 \cdot BC= 3 \cdot YZ\)

\(\displaystyle \frac{ 4 \cdot BC}{4 \cdot YZ}=\frac{ 3 \cdot YZ }{4 \cdot YZ}\)

\(\displaystyle \frac{ BC}{ YZ}=\frac{ 3 }{4 }\)

Again, since only one sidelength ratio is known, similarty can be neither proved nor disproved.

Assume both statements to be true. Similarity, by definition, requires that 

\(\displaystyle \frac{ AB}{ XY} =\frac{ BC}{ YZ}\)

From the two statements together, it can be seen that \(\displaystyle \frac{ AB}{ XY} \ne \frac{ BC}{ YZ}\), so \(\displaystyle \triangle ABC \nsim \triangle XYZ\).

The acute angles of a right triangle are complementary, so \(\displaystyle \angle A\) and \(\displaystyle \angle B\) are a complementary pair, as are \(\displaystyle \angle X\) and \(\displaystyle \angle Y\).

If Statement 1 is assumed—that is, if \(\displaystyle \angle A\) and \(\displaystyle \angle Y\) are a complementary pair—then, since two angles complementary to the same angle—here, \(\displaystyle \angle Y\)—must be congruent, \(\displaystyle \angle A \cong \angle X\). Since right angles \(\displaystyle \angle C \cong \angle Z\),  \(\displaystyle \triangle ABC \sim \triangle XYZ\) follows by way of the Angle-Angle Similarity Postulate, and Statement 1 turns out to provide sufficient information. By a similar argument, Statement 2 is also sufficient.

Assume Statement 1 alone is true. Then, since \(\displaystyle \angle C \cong \angle Z\), both being right angles, and \(\displaystyle \angle A \cong \angle X\) from Statement 1, \(\displaystyle \triangle ABC \sim \triangle XYZ\) follows by way of the Angle-Angle Similarity Postulate. A similar argument shows Statement 2 also provides sufficient information.

Assume both statements are true.

While in two similar triangles, the ratio of the perimeters, given in Statement 1, is indeed equal to that of the ratios of the lengths of the hypotenuses, given in Statement 2, this is not a sufficient condition for similarity. For example:

Case 1: 

\(\displaystyle AC = 9, BC = 12, AC = 15\)

\(\displaystyle XZ = 6, YZ = 8, XY = 10\)

Case 2:

\(\displaystyle AC = 9, BC = 12, AC = 15\)

\(\displaystyle XZ = 8, YZ = 6, XY = 10\)

In each case, the conditions of the main problem and both statements are met, since:

Both triangles are right - each Pythagorean triple is a multiple of Pythagorean triple 3-4-5;

The ratio of the perimeters is \(\displaystyle \frac{9+12+15}{6+8+10}= \frac{36}{24} = \frac{3}{2}\); and,

\(\displaystyle \frac{AC}{XY} = \frac{15}{10}\).

But in Case 1, 

\(\displaystyle \frac{AC}{XZ} = \frac{BC}{YZ} = \frac{AB}{XY}\), since \(\displaystyle \frac{9}{6} = \frac{12}{8} = \frac{15}{10} = \frac{3}{2}\), and the similarity follows by way of the Side-Side-Side Similarity Principle. 

In Case 2, 

\(\displaystyle \frac{AC}{XZ} \ne \frac{AB}{XY}\), since \(\displaystyle \frac{9}{8} \ne \frac{15}{10}\). This violates the conditions of similarity (note that in both cases, \(\displaystyle \triangle ABC \sim \triangle YXZ\), but this is a different statement).

The two statements together are inconclusive.

Each of Statement 1 and Statement 2 gives information about only one of the triangles, so neither statement alone is sufficient.

Assume both statements are true. From Statement 1, \(\displaystyle m \angle A = 45^{\circ }\) and \(\displaystyle \angle B\) is right and measures \(\displaystyle 90^{\circ }\).

From Statement 2 alone, \(\displaystyle \bigtriangleup XYZ\) is isosceles; the acute angles of an isosceles right triangle must both measure \(\displaystyle 45^{\circ }\), so, in particular,  \(\displaystyle m \angle X = 45^{\circ }\). Also, it is given that \(\displaystyle \angle Y\) is right.

\(\displaystyle \angle A \cong \angle X\) and \(\displaystyle \angle B \cong \angle Y\) (both of the latter being right angles), and by the Angle-Angle Postulate, \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup XYZ\).