Answer
$$T_{2}(x) =2-3 x+\frac{5}{2} x^{2}$$
$$T_{3}(x) =2-3 x+\frac{5}{2} x^{2}-\frac{3}{2} x^{3}$$
Work Step by Step
Given $$f(x)=e^{-x}+e^{-2 x}, \quad a=0$$
Since
\begin{aligned}
f(x) &=e^{-x}+e^{-2 x} & & f(0) &=2 \\
f^{\prime}(x) &=-e^{-x}-2 e^{-2 x} & & f^{\prime}(0) &=-3 \\
f^{\prime \prime}(x) &=e^{-x}+4 e^{-2 x} & & f^{\prime \prime}(0) &=5 \\
f^{\prime \prime \prime}(x) &=-e^{-x}-8 e^{-2 x} & & f^{\prime \prime \prime}(0) &=-9
\end{aligned}
then
\begin{aligned}
T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2} \\
&=2+(-3)(x-0)+\frac{5}{2}(x-0)^{2}\\
&=2-3 x+\frac{5}{2} x^{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} \\
&=2+(-3)(x-0)+\frac{5}{2}(x-0)^{2}+\frac{-9}{6}(x-0)^{3}\\
&=2-3 x+\frac{5}{2} x^{2}-\frac{3}{2} x^{3}
\end{aligned}
