= Solution
The <normal equations for linear least squares> are $X^TX\widehat\theta=X^TY$. Under full column rank, $\widehat\theta=(X^TX)^{-1}X^TY$. Since $\operatorname{Cov}(Y)=\sigma^2I$,
$$
\boxed{\operatorname{Cov}(\widehat\theta)=\sigma^2(X^TX)^{-1}X^TX(X^TX)^{-1}=\sigma^2(X^TX)^{-1}}.
$$
Here $\bar x=0$ and $\bar z=2$, so the design matrix is
$$
\boxed{X=\begin{pmatrix}1&-3&1\\1&-2&0\\1&-1&-1\\1&1&-1\\1&2&0\\1&3&1\end{pmatrix}}.
$$
Its columns are pairwise orthogonal and have squared lengths $6,28,4$. Consequently $X^TX=\operatorname{diag}(6,28,4)$, giving
$$
\boxed{\operatorname{Var}(\widehat\beta)=\frac{\sigma^2}{28},\quad\operatorname{Var}(\widehat\gamma)=\frac{\sigma^2}4,\quad\operatorname{Cov}(\widehat\beta,\widehat\gamma)=0}.
$$
Both estimators are linear functions of the <Jointly Gaussian> observation vector, so they are <Jointly Gaussian>. Zero covariance then implies \b[<independence>]. This final implication uses the normal error model, whereas the <variance> calculation itself needs only its second moments.
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