Answer
\begin{aligned}
T_{2}(x) &=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2} \\
T_{3}(x) &= \frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2}-\frac{1}{81}(x-2)^{3}
\end{aligned}
Work Step by Step
Since
\begin{array}{ll}
{f(x)=\frac{1}{1+x},} & {f(2)=\frac{1}{3}} \\
{f^{\prime}(x)=-\frac{1}{(1+x)^{2}},} & {f^{\prime}(2)=-\frac{1}{9}} \\
{f^{\prime \prime}(x)=\frac{2}{(1+x)^{3}},} & {f^{\prime \prime}(2)=\frac{2}{27}} \\
{f^{\prime \prime \prime}(x)=-\frac{6}{(1+x)^{4}},} & {f^{\prime \prime \prime}(2)=-\frac{2}{27}}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2} \\
T_{3}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\
&=\frac{1}{3}-\frac{1}{9}(x-2)+\frac{1}{27}(x-2)^{2}-\frac{1}{81}(x-2)^{3}
\end{aligned}
