The sequence is formed by alternately adding $\displaystyle 4$ and adding $\displaystyle 5$ to each term to get the next term.

$\displaystyle \begin{matrix} 10 + 4 = 14\\ 14 + 5 = 19\\ 19+4= 23\\ 23+5=28\\ 28+4 = 32\\ 32+5=37\\ 37+4=41\\ 41+5=46 \end{matrix}$

$\displaystyle 41$ and $\displaystyle 46$ are the next two numbers.

We demonstrate that all of the choices are subsets of $\displaystyle A \cap B$.

$\displaystyle A \cap B$ is the intersection of $\displaystyle A$ and $\displaystyle B$ - that is, the set of all elements of both sets. Therefore, 

$\displaystyle A \cap B = \left \{ c,e,f,h,j,p\right \}$

$\displaystyle \left \{ c,e,f,h,j,p\right \}$ itself is one of the choices; it is a subset of itself. The empty set $\displaystyle \varnothing$ is a subset of every set. The other two sets listed comprise only elements from $\displaystyle A \cap B$, making them subsets of $\displaystyle A \cap B$.

We are looking for an element that is in the intersection of $\displaystyle A$ and $\displaystyle B$ - in other words, we are looking for an element that appears in both sets.

$\displaystyle A$ is the set of all multiples of 8. We can eliminate two choices as not being in $\displaystyle A$ by demonstrating that dividing each by 8 yields a remainder:

$\displaystyle 484 \div 8 = 60 \textrm{ R }4$

$\displaystyle 900 \div 8 = 112 \textrm{ R }4$

$\displaystyle B$ is the set of all perfect square integers.  We can eliminate two additional choices as not being perfect squares by showing that each is between two consecutive perfect squares:

$\displaystyle 18^{2}= 324< 352 < 361 = 19^{2}$

$\displaystyle 18^{2}= 324< 336 < 361 = 19^{2}$

This eliminates 352 and 336. However, 

$\displaystyle 256 = 16 ^{2}$.

It is also a multiple of 8:

$\displaystyle 256 \div 8 = 32$

Therefore, $\displaystyle 256 \in A \cap B$.

$\displaystyle A \cup B$ is the union of $\displaystyle A$ and $\displaystyle B$, the set of all elements that appear in either set. Therefore, we are looking to eliminate the elements in $\displaystyle A$ and those in $\displaystyle B$ to find the element in neither set.

$\displaystyle A$ is the set of all multiples of 8. We can eliminate two choices as mulitples of 8:

$\displaystyle 272 \div 8 = 34$, so $\displaystyle 272 \in A$

$\displaystyle 344\div 8 = 43$, so $\displaystyle 344 \in A$

$\displaystyle B$ is the set of all perfect square integers. We can eliminate two additional choices as perfect squares:

$\displaystyle \sqrt{225}=15$, so $\displaystyle 225 \in B$

$\displaystyle \sqrt{441}=21$, so $\displaystyle 441 \in B$

All four of the above are therefore elements of $\displaystyle A \cup B$.

420, however is in neither set:

$\displaystyle 420 \div 8 = 52 \textrm{ R } 4$, so $\displaystyle 420 \notin A$

and 

$\displaystyle 20 = \sqrt{400}< \sqrt{420} < \sqrt{441} = 21$, so $\displaystyle 420 \notin B$

Therefore,  $\displaystyle 420 \notin A \cup B$, making this the correct choice.

Derreck is choosing three students from a field of six (seven minus Anne) without respect to order, making this a combination. He has $\displaystyle C(6,3)$ ways to choose. This is:

$\displaystyle C(6,3)= \frac{6!}{(6-3)! \; 3! } = \frac{6!}{3! \; 3! } = \frac{720 }{6 \times 6}= 20$ 

Derreck has 20 ways to fill the ballot.

Four candidates are being selected from ten, with order being important; this means that we are looking for the number of permutations of four chosen from a set of ten. This is

$\displaystyle P (10,4) = \frac{10!}{(10-4)!}= \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5,040$

There are 5,040 ways to complete the ballot.

These are four independent events, so by the multiplication principle, the ballot can be filled out $\displaystyle 4 \times 5 \times 4\times 7 = 560$ ways.

These are three independent events, so by the multiplication principle, the ballot can be filled out $\displaystyle 6 \times 5 \times 6 = 180$ ways.

Since Mike is already chosen, Roy is in essence choosing four candidates from nine, with order being important. This is a permutation of four elements out of nine. The number of these is

$\displaystyle P (9,4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3,024$

Roy can fill out the ballot 3,024 times and have Mike be his first choice.

To find the next number in the list, multiply the previous number by $\displaystyle \small 3$.

$\displaystyle 27\times3=81$