= Solution
Put $\mathrm{RSS}_1=\sum_i(Y_i-\widehat\alpha-\widehat\beta x_i)^2$ for the unrestricted fit. Under $H_0$, the best slope is still $\widehat\beta$ because the design is centered. Orthogonality of the residuals to the constant vector gives
$$
\mathrm{RSS}_0=\sum_i(Y_i-\widehat\beta x_i)^2=\mathrm{RSS}_1+n\overline Y^2.
$$
At either fitted mean, the <maximum-likelihood estimator> of variance is $\mathrm{RSS}/n$. The ratio used by the <generalized likelihood-ratio test> is consequently
$$
\Lambda=\left(\frac{\mathrm{RSS}_1}{\mathrm{RSS}_0}\right)^{n/2}=\left(1+\frac{T_n^2}{n-2}\right)^{-n/2},\qquad T_n=\frac{\sqrt n\,\overline Y}{\sqrt{\mathrm{RSS}_1/(n-2)}}.
$$
Small likelihood ratios correspond to large $|T_n|$. Under $H_0$, part (a) gives independent $\sqrt n\,\overline Y/\sigma\sim N(0,1)$ and $\mathrm{RSS}_1/\sigma^2\sim\chi^2_{n-2}$. Thus $T_n$ has the <Student's t-distribution> with $n-2$ degrees of freedom. \b[The exact size-5% critical region is]
$$
\boxed{|T_n|>t_{n-2,\,0.975},}
$$
where $t_{\nu,0.975}$ is the 97.5th percentile of that distribution. The denominator uses the unbiased residual variance estimate, not the likelihood variance denominator $n$.
Under an alternative, the same statistic has the representation
$$
T_n=\frac{Z+\sqrt n\,\alpha/\sigma}{\sqrt{W/(n-2)}},\qquad Z\sim N(0,1),\quad W\sim\chi^2_{n-2},\quad Z\perp W.
$$
The joint law of $Z,W$ is parameter-free, and $\beta$ has disappeared. Therefore
$$
\boxed{w(\alpha,\beta,\sigma^2)=g(\alpha/\sigma),}
$$
where $g(d)=\mathbb P(|(Z+\sqrt n d)/\sqrt{W/(n-2)}|>t_{n-2,0.975})$. This is also the <noncentral t-distribution> description of the power. As explained in part (a), an exact residual-based test requires $n>2$.
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