= Solution
The <Neyman-Pearson lemma> concerns a <simple hypothesis> with density $f_0$ against a simple alternative with density $f_1$. Allow a randomized test $0\le\varphi\le1$. Choose a threshold $c\ge0$ and, if necessary, randomize on the equality set so that
$$
\varphi_*=1\quad\text{where }f_1>cf_0,\qquad
\varphi_*=0\quad\text{where }f_1<cf_0,\qquad E_0\varphi_*=\alpha.
$$
Then $\varphi_*$ is a <most powerful test> among all tests of size at most $\alpha$. To prove it, let $E_0\varphi\le\alpha$. Pointwise, $(\varphi_*-\varphi)(f_1-cf_0)\ge0$: above threshold both factors are nonnegative; below threshold both are nonpositive; on the threshold their product vanishes. Therefore
$$
E_1\varphi_*-E_1\varphi
\ge c(E_0\varphi_*-E_0\varphi)\ge0.
$$
This proves the lemma, including regions with $f_0=0$, where a positive alternative density must be rejected. It also shows that ties in the <likelihood ratio> are the only discretionary region when the inequality is strict off the threshold.
For the stated family, $R=|X|$ has the <Gamma distribution> with shape $\theta$ and unit rate. At $\theta=1$ it has the <exponential distribution>, and the <likelihood ratio> for $\theta=2$ against $\theta=1$ is
$$
\frac{f(x\mid2)}{f(x\mid1)}=|x|,
$$
because $\Gamma(1)=\Gamma(2)=1$. The <most powerful test> therefore rejects for large $|X|$. Since $\Pr_1(|X|>c)=e^{-c}$, its exact size-$\alpha$ critical value and rejection rule are
$$
\boxed{c=-\log\alpha,\qquad\text{reject }H_0\text{ if }|X|>-\log\alpha.}
$$
The distribution is <continuous>, so boundary randomization is unnecessary. Under $\theta=2$, <integration by parts> gives the <power function of a statistical test>
$$
\boxed{\Pr_2(|X|>c)=\int_c^\infty r e^{-r}\,dr
=(c+1)e^{-c}=\alpha(1-\log\alpha).}
$$
For every fixed $\theta>1$, the <likelihood ratio> against the same null is $|x|^{\theta-1}/\Gamma(\theta)$, strictly increasing in $|x|$. The size constraint still selects exactly the same critical value $c=-\log\alpha$, independent of the alternative. The <Neyman-Pearson lemma> therefore makes this single test most powerful against each $\theta>1$ separately. Hence \b[it is uniformly most powerful for the composite alternative $\theta>1$].
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