The \(\displaystyle x^{2}\)  term of the Maclaurin series of a function \(\displaystyle f\) has coefficient \(\displaystyle \frac{f '' (0) }{2!} = \frac{f '' (0) }{2}\)

The second derivative of \(\displaystyle f\) can be found as follows:

\(\displaystyle f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}\)

\(\displaystyle f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}\)

\(\displaystyle f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}\)

\(\displaystyle f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}\)

The coeficient of \(\displaystyle x^{2}\) in the Maclaurin series is therefore

\(\displaystyle \frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}\)

The \(\displaystyle (x-1)^{2}\) term of a Taylor series expansion about \(\displaystyle x = 1\) is

\(\displaystyle \frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}\).

We can find \(\displaystyle f''(x)\) by differentiating twice in succession:

\(\displaystyle f (x) = \log_5 x\)

\(\displaystyle f'(x) = \frac{1}{x \ln 5}\)

\(\displaystyle f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}\)

\(\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )\)

\(\displaystyle f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )\)

\(\displaystyle f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )\)

\(\displaystyle f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}\)

\(\displaystyle f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }\)

so the \(\displaystyle (x-1)^2\) term is \(\displaystyle \frac{f''(1) }{2} \; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}\)

This can most easily be answered by recalling that the Maclaurin series for \(\displaystyle \tan^{-1} x\) is 

\(\displaystyle \tan^{-1} x = x - \frac{1}{3} x^{3 } + \frac{1}{5} x^{5 }...\)

Multiply by \(\displaystyle x\) to get:

 \(\displaystyle x\tan^{-1} x = x\cdot x - x\cdot \frac{1}{3} x^{3 } + x\cdot \frac{1}{5} x^{5 }...\)

\(\displaystyle x \tan^{-1} x=x^{2 } - \frac{1}{3} x^{4 } + \frac{1}{5} x^{6 }...\)

The \(\displaystyle x^{4}\) term is therefore \(\displaystyle - \frac{1}{3} x^{4 }\).

The \(\displaystyle x^ {3}\) term of a Maclaurin series expansion has coefficient

\(\displaystyle \frac{f ' '' (0)}{3!} = \frac{f ' '' (0)}{6}\).

We can find \(\displaystyle f'''(x)\) by differentiating three times in succession:

\(\displaystyle f(x) = \sinh 3x\)

\(\displaystyle f'(x) = 3 \cosh 3x\)

\(\displaystyle f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x\)

\(\displaystyle f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x\)

\(\displaystyle f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27\)

The term we want is therefore

\(\displaystyle \frac{f ' '' (0)}{6} \cdot x^{3} = \frac{27}{6} \cdot x^{3} = \frac{9}{2} x^{3}\)

The \(\displaystyle x^{2}\) term of a Maclaurin series expansion has coefficient

\(\displaystyle \frac{f ' ' (0)}{2!} = \frac{f ' ' (0)}{2}\).

We can find \(\displaystyle f''(x)\) by differentiating twice in succession:

\(\displaystyle f (x) = 7^{x}\)

\(\displaystyle f ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )= \ln 7 \cdot 7^{x}\)

\(\displaystyle f ' '(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left (\ln 7 \cdot 7^{x} \right )\)

\(\displaystyle f ' '(x)= \ln 7\cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )\)

\(\displaystyle f ' '(x)= \ln 7\cdot \ln 7\cdot 7^{x}\)

\(\displaystyle f ' '(x)=\left ( \ln 7 \right )^{2} \cdot 7^{x}\)

\(\displaystyle f ' '(0)=\left ( \ln 7 \right )^{2} \cdot 7^{0} = \left ( \ln 7 \right )^{2} \cdot 1 = \left ( \ln 7 \right )^{2}\)

The coefficient we want is 

\(\displaystyle \frac{f ' ' (0)}{2} =\frac{\left ( \ln 7 \right )^{2}}{2}\),

so the corresponding term is \(\displaystyle \frac{\left ( \ln 7 \right )^{2}}{2} x^{2}\).