Answer
$$T_{2}(x) =4+\frac{8}{1 !}(x-\ln (2))+\frac{16}{2 !}(x-\ln (2))^{2} $$
$$T_{3}(x) =4+8(x-\ln 2)+\frac{16}{2!}(x-\ln 2)^{2}+\frac{32}{3!}(x-\ln 2)^{3}$$
Work Step by Step
Given $$f(x)=e^{2 x}, \quad a=\ln 2$$
Since
\begin{array}{rlrl}
f(x) & =e^{2 x} & f(\ln 2)&=4 \\
f^{\prime}(x) & =2e^{2 x} & f^{\prime}(\ln 2) & =8 \\
f^{\prime \prime}(x) & =4e^{2 x}& f^{\prime \prime}(\ln 2) & =16 \\
f^{\prime \prime \prime}(x) & =8e^{2 x} & f^{\prime \prime \prime}(\ln 2) & =32
\end{array}
Then
\begin{align*}
T_{2}(x)&=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}\\
&=4+\frac{8}{1 !}(x-\ln (2))+\frac{16}{2 !}(x-\ln (2))^{2}
\end{align*}
and
\begin{aligned}
T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{6}(x-a)^{3} \\
&=4+8(x-\ln 2)+\frac{16}{2!}(x-\ln 2)^{2}+\frac{32}{3!}(x-\ln 2)^{3}
\end{aligned}
