5. Given $\boldsymbol{a}=(\cos \alpha, \sin \alpha), \boldsymbol{b}=(\cos \beta, \sin \beta)$, $|a-b|=\frac{2 \sqrt{5}}{5}$. If $0<\alpha<\frac{\pi}{2},-\frac{\pi}{2}<\beta<0$, and $\sin \beta=-\frac{5}{13}$, then $\sin \alpha=(\quad)$.
(A) $\frac{31}{65}$
(B) $\frac{32}{65}$
(C) $\frac{33}{65}$
(D) $\frac{34}{65}$
5.C.
$$
\begin{array}{l}
|a-b|^{2}=(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2} \\
=2-2(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta) \\
=2-2 \cos (\alpha-\beta)
\end{array}
$$

That is, $2-2 \cos (\alpha-\beta)=\frac{4}{5}, \cos (\alpha-\beta)=\frac{3}{5}$.
Given $\alpha=(\alpha-\beta)+\beta$, and $0<\alpha<\frac{\pi}{2},-\frac{\pi}{2}<$
$\beta<0$ and $\sin \beta=-\frac{5}{13}, \cos (\alpha-\beta)=\frac{3}{5}$, we know $\cos \beta=\frac{12}{13}, \sin (\alpha-\beta)=\frac{4}{5}$.
Therefore, $\sin \alpha=\sin [\beta+(\alpha-\beta)]$
$$
=\sin \beta \cdot \cos (\alpha-\beta)+\cos \beta \cdot \sin (\alpha-\beta)=\frac{33}{65} .
$$