= Solution
Assume the <variances> are finite. Expand the square and apply <linearity of expectation>:
$$
\begin{aligned}
\operatorname{var}(X+Y)
&=\mathbb E[X^2]+2\mathbb E[XY]+\mathbb E[Y^2]-(\mathbb E[X]+\mathbb E[Y])^2\\
&=\operatorname{var}(X)+\operatorname{var}(Y)+2\bigl(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]\bigr).
\end{aligned}
$$
Part (i), applied to the identity <functions>, makes the bracketed <covariance> zero by <independence of random variables>. Hence
$$
\boxed{\operatorname{var}(X+Y)=\operatorname{var}(X)+\operatorname{var}(Y).}
$$
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