= Solution
By <linearity of expectation>, \b[$\mathbb E[Z]=\sum_i\mathbb E[X_i]=0$]. Expanding the square and again using <linearity of expectation> gives
$$
\begin{aligned}
\operatorname{Var}(Z)
&=\mathbb E[Z^2]
=\sum_i\mathbb E[X_i^2]+2\sum_{i<j}\mathbb E[X_iX_j]\\
&=\boxed{\sum_i\sigma_i^2+2\sum_{i<j}\operatorname{Cov}(X_i,X_j)}.
\end{aligned}
$$
Here the zero means imply $\mathbb E[X_i^2]=\sigma_i^2$ and $\mathbb E[X_iX_j]=\operatorname{Cov}(X_i,X_j)$. The <covariance> expansion itself does not require <independence>. Under the additional <independence> stated here, $\mathbb E[X_iX_j]=\mathbb E[X_i]\mathbb E[X_j]=0$, so it simplifies further to \b[$\operatorname{Var}(Z)=\sum_i\sigma_i^2$].
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