Answer
\begin{aligned}
T_{2} &=-5-(x+2)-2(x+2)^{2}\\
T_{3} &=-5-(x+2)-2(x+2)^{2}-2(x+2)^{3}
\end{aligned}
Work Step by Step
Since
\begin{array}{ll}
{f(x)=\frac{x^2+1}{x+1},} & {f(-2)=-5} \\
{f^{\prime}(x)=\frac{x^2+2x-1}{\left(x+1\right)^2},} & {f^{\prime}(-2)=-1} \\
{f^{\prime \prime}(x)=\frac{4}{\left(x+1\right)^3},} & {f^{\prime \prime}(-2)=-4} \\
{f^{\prime \prime \prime}(x)=-\frac{12}{\left(x+1\right)^4},} & {f^{\prime \prime \prime}(-2)=-12}
\end{array}
Then
\begin{aligned}
T_{2}&=-5+\frac{-1}{1 !}(x+2)+\frac{-4}{2 !}(x+2)^{2}\\
&=-5-(x+2)-2(x+2)^{2}\\
T_{3}&=-5+\frac{-1}{1 !}(x+2)+\frac{-4}{2 !}(x+2)^{2}+\frac{-12}{3 !}(x+2)^{3}\\
&=-5-(x+2)-2(x+2)^{2}-2(x+2)^{3}
\end{aligned}
