(a)
Since \(X\) and \(\epsilon\) are independent, we have \(E_{\mathcal{T}}(\epsilon) = 0\). Since \(\hat \beta = (\textbf{X}^T\textbf{X})^{-1}\textbf{X}^Ty = \beta + (\textbf{X}^T\textbf{X})^{-1}\textbf{X}^T\epsilon\), we have \(E_{\mathcal{T}}(\hat \beta) = \beta\) and
\[\begin{equation}
\label{eq:2-5a}
\text{Var}_\mathcal{T}(\hat\beta) = E_\mathcal{T}(\hat \beta^T\hat\beta) = E_\mathcal{T}(\textbf{X}^T\textbf{X})^{-1}\sigma^2.\ \ \ \ \ \ \
\end{equation}\]
Note that \(y_0\) is constant for the distribution \(\mathcal{T}\). We have
\[\begin{eqnarray}
E_\mathcal{T}(y_0-\hat y_0)^2 &=& y_0^2 + E_\mathcal{T}\hat y_0^2 - 2 y_0E_\mathcal{T}\hat y_0\nonumber\\
&=& [y_0^2-E_{y_0|x_0}(y_0)^2] + [E_\mathcal{T}\hat y_0^2 - (E_\mathcal{T}\hat y_0)^2]\nonumber\\
&& + [ (E_\mathcal{T}\hat y_0)^2 - 2 y_0E_\mathcal{T}\hat y_0 + E_{y_0|x_0}(y_0)^2 ]\nonumber\\
&=& [y_0^2-E_{y_0|x_0}(y_0)^2] + [E_\mathcal{T}\hat y_0^2 - (E_\mathcal{T}\hat y_0)^2]\nonumber\\
&& + (E_\mathcal{T}\hat y_0 - y_0)^2\nonumber
\end{eqnarray}\]
Therefore, by \(\eqref{eq:2-5a}\), we have
\[\begin{eqnarray}
E_{y_0|x_0}E_\mathcal{T}(y_0-\hat y_0)^2 &=&\text{Var}(y_0|x_0) + \text{Var}_\mathcal{T}(\hat y_0) + \text{Bias}^2(\hat y_0)\nonumber\\
&=&\sigma^2 + E_\mathcal{T} x_0^T(\textbf{X}^T\textbf{X})^{-1}x_0\sigma^2 + 0^2.\nonumber
\end{eqnarray}\]
(b)
First, we have
\[\begin{equation}
E_{x_0}\text{EPE}(x_0) \sim E_{x_0}x_0^T\text{Cov}(X)^{-1}x_0\sigma^2/N + \sigma^2.\nonumber
\end{equation}\]
Note that \(x_0^T\text{Cov}(X)^{-1}x_0\) is scalar and equal to its own trace, we have
\[\begin{eqnarray}
E_{x_0}\text{EPE}(x_0) &\sim& E_{x_0}x_0^T\text{Cov}(X)^{-1}x_0\sigma^2/N + \sigma^2\nonumber\\
&=&\text{trace}\left(E_{x_0}x_0x_0^T\text{Cov}(X)^{-1}\right)\sigma^2/N + \sigma^2\nonumber\\
&=&\text{trace}\left(\textbf{I}_p\right)\sigma^2/N + \sigma^2\nonumber\\
&=&\sigma^2(p/N) + \sigma^2.\nonumber
\end{eqnarray}\]