As $X$ and $Y$ are standard normal variate, therefore: $$X^2 \sim \chi^2(1)\hspace{.5cm} \text{let's call it}\hspace{.5cm}\chi_X^2(1)$$ $$\text&$$ $$Y^2 \sim \chi^2(1)\hspace{.5cm} \text{let's call it}\hspace{.5cm}\chi_Y^2(1)$$
As $X$ and $Y$ are independent, $X^2$ and $Y^2$ are also independent. Thus, $\chi_X^2(1)$ and $\chi_Y^2(1)$ are independent.
We also know that if $X$ and $Y$ are two independent random variables and $M(t)$ denotes the moment generating function of a given random variable then: $$M_{aX+bY}(t)=M_X(at)\cdot M_Y(bt)$$
As the MGF of $\chi^2(1)$ is $M(t)=\frac{1}{\sqrt{1-2t}}$, we have
$$M_{\frac{\chi_X^2(1)}{2}+\frac{-\chi_Y^2(1)}{2}}(t) = M_{\chi_X^2}\Bigl(\frac{t}{2}\Bigr)\cdot M_{\chi_Y^2}\Bigl(\frac{-t}{2}\Bigr)$$
$$=\frac{1}{\sqrt{1-\frac{2t}{2}}}\cdot\frac{1}{\sqrt{1+\frac{2t}{2}}}$$ $$=\frac{1}{\sqrt{1-t^2}}$$
which is not MGF of any distribution known to me. But the answer says it's actually $\frac{X^2-Y^2}{2} \sim exp(\lambda=1)$
Moreover, I am getting $\frac{X^2+Y^2}{2} \sim exp(\lambda=1)$ instead. As
$$M_{\frac{\chi_X^2(1)}{2}+\frac{\chi_Y^2(1)}{2}}(t) = M_{\chi_X^2}\Bigl(\frac{t}{2}\Bigr)\cdot M_{\chi_Y^2}\Bigl(\frac{t}{2}\Bigr)$$
$$=\frac{1}{\sqrt{1-\frac{2t}{2}}}\cdot\frac{1}{\sqrt{1-\frac{2t}{2}}}$$ $$=\frac{1}{\sqrt{1-t}^2}=\frac{1}{1-t}$$ which is MGF of $exp(\lambda=1)$
Am I doing something wrong or there is a misprint in question?