= Solution
A <simple hypothesis> specifies the entire probability distribution. A <critical region> $C$ is the set of observations for which the <null hypothesis> is rejected. The <size of a statistical test> is $\sup_{\theta\in H_0}\Pr_\theta(X\in C)$, which reduces to $\Pr_0(C)$ for a <simple hypothesis>. The <power function> is $\Pr_\theta(C)$ as a function of the true parameter or distribution. At a specified alternative, the <Type II error> probability is $\Pr_\theta(C^c)=1-\Pr_\theta(C)$, the chance of failing to reject a false <null hypothesis>. The <Type I error> probability is the corresponding rejection probability under the <null hypothesis>.
The <Neyman-Pearson lemma> says that, for two <simple hypotheses>, a test rejecting where $f_1>kf_0$, accepting where $f_1<kf_0$, and if necessary randomizing on equality to achieve prescribed size $\alpha$, has maximal <statistical power> among tests of size at most $\alpha$. Conversely a most powerful test can be taken in this <likelihood ratio> form, up to equality sets and null sets.
Here the <likelihood ratio> for $x\ne0$ is
$$
\frac{f_1(x)}{f_0(x)}=\sqrt{\frac2\pi}\frac1{|x|}.
$$
Its value is infinite at $0$, which has probability zero under both continuous distributions. Large <likelihood ratios> correspond to small $|x|$. Thus the best <critical region> is $|X|<c_\alpha$, where
$$
\alpha=\int_{-c_\alpha}^{c_\alpha}\frac12|x|e^{-x^2/2}\,dx
=1-e^{-c_\alpha^2/2}.
$$
There is no boundary randomization to worry about. The complete answer is
$$
\boxed{\text{reject if }|X|<\sqrt{-2\log(1-\alpha)},\qquad
\text{power}=2\Phi\!\left(\sqrt{-2\log(1-\alpha)}\right)-1}.
$$
For the <minimum sum of errors in a simple hypothesis test>, write the error sum as $\int_Cf_0+\int_{C^c}f_1$. At each observation its smaller contribution is obtained by rejecting exactly where $f_1>f_0$. This proves optimality over all tests, including randomized ones, rather than just optimizing within an assumed family. Here it gives $c_*=\sqrt{2/\pi}$, and therefore
$$
\boxed{\alpha_*=1-e^{-1/\pi}\approx0.2726}.
$$
The minimum error sum is $1-e^{-1/\pi}+2[1-\Phi(\sqrt{2/\pi})]$. This criterion weights the two error probabilities equally; it does not impose a conventional $5\%$ <significance level>.
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