Answer
\[\begin{align}
& \left. a \right)2+2\left( x-\ln 2 \right)+{{\left( x-\ln 2 \right)}^{2}}+\frac{1}{3}{{\left( x-\ln 2 \right)}^{3}} \\
& \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{2}{k!}}{{\left( x-\ln 2 \right)}^{k}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{e}^{x}},\text{ }a=\ln 2 \\
& \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\
& \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ }\left( \mathbf{1} \right) \\
& =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \\
& \text{Calculate some derivatives and their value at }x=\ln2 \\
& f\left( x \right)={{e}^{x}}\to f\left( \ln 2 \right)={{e}^{\ln 2}}=2 \\
& f'\left( x \right)={{e}^{x}}\to f\left( \ln 2 \right)={{e}^{\ln 2}}=2 \\
& f''\left( x \right)={{e}^{x}}\to f\left( \ln 2 \right)={{e}^{\ln 2}}=2 \\
& f'''\left( x \right)={{e}^{x}}\to f\left( \ln 2 \right)={{e}^{\ln 2}}=2 \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =\frac{2}{0!}{{\left( x-\ln 2 \right)}^{0}}+\frac{2}{1!}{{\left( x-\ln 2 \right)}^{1}}+\frac{2}{2!}{{\left( x-\ln 2 \right)}^{2}}+\frac{2}{3!}{{\left( x-\ln 2 \right)}^{3}}\left( \mathbf{2} \right) \\
& \\
& \text{Simplifying} \\
& =2+2\left( x-\ln 2 \right)+{{\left( x-\ln 2 \right)}^{2}}+\frac{1}{3}{{\left( x-\ln 2 \right)}^{3}} \\
& \\
& \left. b \right)\text{ Taking the result }\left( \mathbf{2} \right) \\
& =\frac{2}{0!}{{\left( x-\ln 2 \right)}^{0}}+\frac{2}{1!}{{\left( x-\ln 2 \right)}^{1}}+\frac{2}{2!}{{\left( x-\ln 2 \right)}^{2}}+\frac{2}{3!}{{\left( x-\ln 2 \right)}^{3}} \\
& \text{Using summation notation}\text{, we obtain} \\
& =\sum\limits_{k=0}^{\infty }{\frac{2}{k!}}{{\left( x-\ln 2 \right)}^{k}} \\
\end{align}\]
