= Solution
A <simple hypothesis> specifies a single probability distribution. For a test with rejection region $R$, its <size of a statistical test> is the rejection probability under the simple <null hypothesis>, while its <statistical power> against the simple alternative is the rejection probability under that alternative. The <Neyman-Pearson lemma> states that among tests of size at most $\alpha$, a test rejecting for the largest values of the <likelihood ratio> $f_1/f_0$ has greatest power, with boundary randomization if needed.
Here
$$
\frac{f_1(x)}{f_0(x)}=\frac32(1-x^2),
$$
which decreases with $|x|$. The best test therefore rejects for $|X|\leq c$. Under $H_0$, $\mathbb P_0(|X|\leq c)=c$, so size $0.05$ requires $c=0.05=1/20$. Its power is
$$
\boxed{\mathbb P_1(|X|\leq1/20)
=\int_{-1/20}^{1/20}\frac34(1-x^2)\,dx
=\frac{1199}{16000}=0.0749375.}
$$
\b[Thus reject $H_0$ exactly when $|X|\leq0.05$.]
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