= Solution
Let $Q=\sum_i x_i^2>0$. Under the usual nondegenerate regression assumption $n>2$, maximizing the normal likelihood is equivalent to minimizing the <residual sum of squares>. The normal equations and $\sum_i x_i=0$ give the <maximum-likelihood estimators>
$$
\boxed{\widehat\alpha=\overline Y,\qquad\widehat\beta=\frac{\sum_i x_iY_i}{Q}.}
$$
Write $A=\overline\varepsilon$ and $B=\sum_i x_i\varepsilon_i/Q$. Then $\widehat\alpha=\alpha+A$, $\widehat\beta=\beta+B$, and the residual is $r_i=\varepsilon_i-A-x_iB$. Independence of the errors gives
$$
\operatorname{Var}(A)=\sigma^2/n,\quad\operatorname{Var}(B)=\sigma^2/Q,\quad\operatorname{Cov}(A,B)=\frac{\sigma^2\sum_i x_i}{nQ}=0.
$$
For each $i$,
$$
\operatorname{Cov}(A,r_i)=\sigma^2/n-\sigma^2/n=0,\qquad\operatorname{Cov}(B,r_i)=\sigma^2x_i/Q-x_i\sigma^2/Q=0.
$$
Thus the three requested variables are pairwise uncorrelated.
They are linear functions of the jointly normal error vector. More strongly, the whole vector $(A,B,r_1,\ldots,r_n)$ has a <multivariate normal distribution>, and its block covariances show that $A$, $B$ and the residual vector are independent, by <uncorrelated jointly normal variables are independent>. Independence is preserved when applying a measurable function to the residual block. Hence
$$
\boxed{\widehat\alpha,\ \widehat\beta,\ \sum_i r_i^2\text{ are mutually independent}.}
$$
The residual projection is $I-\mathbf1\mathbf1^{\mathsf T}/n-xx^{\mathsf T}/Q$, of rank $n-2$, so additionally $\sum_i r_i^2/\sigma^2\sim\chi^2_{n-2}$. This is the <orthogonal projection of a Gaussian vector> interpretation of the same result. The printed design conditions alone allow $n=2$; then the fit is exact, the positive-variance likelihood has no finite maximizer, and the residual test in part (b) is undefined. The standard testing question therefore needs $n>2$.
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