Answer
$$f(x)=e^{-2}e^{ x}
=\frac{1}{e^2}+\frac{x}{e^2}+\frac{x^2}{2!e^2}+\frac{x^3}{3!e^2}+\frac{x^4}{4!e^2}+...$$
Convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows
$$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
Now by the comparison with the function $f(x)= e^{ x-2}=e^{-2}e^x$, we have the Maclaurin series as follows
$$f(x)=e^{-2}e^{ x}\\
=e^{-2}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\
=\frac{1}{e^2}+\frac{x}{e^2}+\frac{x^2}{2!e^2}+\frac{x^3}{3!e^2}+\frac{x^4}{4!e^2}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)= e^{ x-2}$ is also convergent for any value of $x$.
