Answer
\begin{aligned}
T_{2}(x) &=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2} \\
T_{3}(x)
&=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2}
\end{aligned}
Work Step by Step
Since
\begin{array}{ll}
{f(x)=\frac{1}{1+x^{2}},} & {f(-1)=\frac{1}{2}} \\
{f^{\prime}(x)=-\frac{2 x}{\left(1+x^{2}\right)^{2}},} & {f^{\prime}(-1)=\frac{1}{2}} \\
{f^{\prime \prime}(x)=\frac{2\left(x^{2}-1\right)}{\left(1+x^{2}\right)^{3}},} & {f^{\prime \prime}(-1)=\frac{1}{2}} \\
{f^{\prime \prime \prime}(x)=-\frac{24 x\left(x^{2}-1\right)}{\left(1+x^{2}\right)^{4}},} & {f^{\prime \prime \prime}(-1)=0}
\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}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{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}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2}+(0)(x+1)^{3} \\
&=\frac{1}{2}+\frac{1}{2}(x+1)+\frac{1}{4}(x+1)^{2}
\end{aligned}
