= Solution
The <random variable> $Y=Z^2/2$ is nonnegative, so its <cumulative distribution function> is zero for $y<0$. If $y\ge0$, then
$$
\begin{aligned}
F_Y(y)&=\mathbb P(Z^2/2\le y)
=\mathbb P(-\sqrt{2y}\le Z\le\sqrt{2y})\\
&=\Phi(\sqrt{2y})-\Phi(-\sqrt{2y})
=2\Phi(\sqrt{2y})-1.
\end{aligned}
$$
The last step uses the <symmetry> of the <standard normal distribution>, $\Phi(-z)=1-\Phi(z)$. Hence
$$
\boxed{F_Y(y)=\begin{cases}
2\Phi(\sqrt{2y})-1,&y\ge0,\\
0,&y<0.
\end{cases}}
$$
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