How to prove these two properties? For two independent r.v. $X\sim N(\mu_1, \sigma^2)$ and $Y\sim N(\mu_2, \sigma^2)$. Show that
$$
P(\max\{X^2, Y^2\}>c)\le \alpha.
$$
 A: I assume $\chi_1^2(\alpha)$ is the quantile of a $\chi^2_1$ distribution corresponding to upper tail area $\alpha$, i.e. $P(W>\chi_1^2(\alpha))=\alpha$ for $W\sim \chi^2_1.$
Note that $\mu_1\mu_2=0$ implies at least one of $\mu_1,\mu_2$ is zero. WLOG, assume $\mu_2=0,$ so $Y^2\sim \chi^2_1.$ Then
$$P\left(\min\{X^2,Y^2\}>\chi^2_1(\alpha)\right)=P\left(X^2>\chi^2_1(\alpha),Y^2>\chi^2_1(\alpha)\right)\\
\underbrace{=}_{\text{indep}}\underbrace{P\left(X^2>\chi^2_1(\alpha)\right)}_{\leq 1}\underbrace{P\left(Y^2>\chi^2_1(\alpha)\right)}_{=\alpha}\leq \alpha.$$
Note that $X^2$ is non-central chi squared with one degree of freedom and noncentrality parameter $\mu_1^2.$ This should help you with the second part where they want you to consider the probability as a function of $\mu_1$.

Note: As you have correctly noted in the comments, in this case, you can can use the fact that for $c\geq 0,$
$$P(X^2>c)=P(|X|>\sqrt c)\\=P(X>\sqrt c)+P(X<-\sqrt c)\\=P(X-\mu_1>\sqrt c-\mu_1)+P(X-\mu_1<-\sqrt c-\mu_1)\\=1-\Phi(\sqrt c-\mu_1)+\Phi(-\sqrt c-\mu_1),$$
where $\Phi$ is the CDF of a standard normal.
