= Solution
{c}
Minimizing the squared residual norm gives the <normal equations>
$$
\boxed{X^TX\,\widehat\theta=X^TY.}
$$
Full column rank makes $X^TX$ invertible. With columns corresponding to $1,x,x^2-4$, the <design matrix> is
$$
X=\begin{pmatrix}
1&-3&5\\1&-2&0\\1&-1&-3\\1&0&-4\\
1&1&-3\\1&2&0\\1&3&5
\end{pmatrix}.
$$
Symmetry gives zero cross products between the odd column and either even column, and $\sum(x^2-4)=28-28=0$ gives the remaining orthogonality. Their squared norms are seven, 28 and 84, so $X^TX=\operatorname{diag}(7,28,84)$. Therefore
$$
\boxed{\widehat\alpha=\frac17\sum_{j=1}^7y_j,}
$$
$$
\boxed{\widehat\beta=
\frac{-3y_1-2y_2-y_3+y_5+2y_6+3y_7}{28},}
$$
$$
\boxed{\widehat\gamma=
\frac{5y_1-3y_3-4y_4-3y_5+5y_7}{84}.}
$$
Since the error <covariance> is $\sigma^2I$, <least squares> has <covariance> $\sigma^2(X^TX)^{-1}$. Thus
$$
\boxed{\operatorname{Var}\widehat\alpha=\frac{\sigma^2}7,\quad
\operatorname{Var}\widehat\beta=\frac{\sigma^2}{28},\quad
\operatorname{Var}\widehat\gamma=\frac{\sigma^2}{84}.}
$$
The estimators have zero <covariances>. <independence> would require an additional distributional assumption such as Gaussian errors; the stated mean and <variance> assumptions alone give only the <covariance> calculation.
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