Answer
$$T_{2}(x) =\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2} $$
$$T_{3}(x) =\frac{1}{e}+\frac{1}{e}(x-1)-\frac{1}{2 e}(x-1)^{2}-\frac{1}{6 e}(x-1)^{3}$$
Work Step by Step
Given $$f(x)=x^{2} e^{-x}, \quad a=1$$
Since
\begin{aligned}
f(x) &=x^{2} e^{-x} & & f(1) &=1 / e \\
f^{\prime}(x) &=\left(2 x-x^{2}\right) e^{-x} & & f^{\prime}(1) &=1 / e \\
f^{\prime \prime}(x) &=\left(x^{2}-4 x+2\right) e^{-x} & & f^{\prime \prime}(1) &=-1 / e \\
f^{\prime \prime \prime}(x) &=\left(-x^{2}+6 x-6\right) e^{-x} & & f^{\prime \prime \prime}(1) &=-1 / e
\end{aligned}
Then
\begin{aligned}
T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2} \\
&=\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2}
\end{aligned}
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} \\
&=\frac{1}{e}+\frac{1}{e}(x-1)+\frac{-1 / e}{2}(x-1)^{2}+\left(\frac{-1 / e}{6}\right)(x-1)^{3} \\
&=\frac{1}{e}+\frac{1}{e}(x-1)-\frac{1}{2 e}(x-1)^{2}-\frac{1}{6 e}(x-1)^{3}
\end{aligned}
