= Solution
A <simple hypothesis> specifies a single probability law, with no unknown parameter remaining. For a possibly randomized <hypothesis test>, let $\varphi(x)\in[0,1]$ be its conditional rejection probability. Its <test size> is $E_0\varphi(X)$ and its <statistical power> against the specified alternative is $E_1\varphi(X)$.
The <Neyman-Pearson lemma> states that for two densities $f_0,f_1$ with respect to a common measure, a <likelihood-ratio test> rejecting when $f_1/f_0>k$, accepting when it is below $k$, and randomizing on equality to obtain size $\alpha$, maximizes power among all tests of size at most $\alpha$.
Here the <likelihood ratio> is
$$
\Lambda(x)=\frac{\sqrt{2\pi}}4\exp\left(\frac{x^2-|x|}{2}\right).
$$
Writing $r=|x|$, its logarithm apart from the constant is $(r^2-r)/2$. This decreases until $r=1/2$ and then increases. In particular it is nonpositive for $0\leq r\leq1$ and positive for $r>1$. The supplied normal-tail value and $\alpha<1/4$ imply that the quantile $t$ defined by $2[1-\Phi(t)]=\alpha$ satisfies $t>1$. Hence the threshold $\Lambda(t)$ is above every central value on $r\leq1$; outside that interval the ratio is strictly increasing. The <Neyman-Pearson lemma> therefore selects precisely the two tails:
$$
\boxed{\text{reject }H_0\text{ if }|X|>t,\qquad \Phi(t)=1-\alpha/2}.
$$
The equality boundary has probability zero under both continuous laws, so no randomization is needed. Under the alternative density, the <statistical power> is
$$
\boxed{\int_{|x|>t}\frac14e^{-|x|/2}\,dx=e^{-t/2}}.
$$
The central dip in the <likelihood ratio> is why checking the threshold against the entire central region, rather than assuming monotonicity in $|x|$ everywhere, is necessary.
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