(5) Given $\vec{a}=(\cos \alpha, \sin \alpha), \vec{b}=(\cos \beta, \sin \beta),|\vec{a}-\vec{b}|=\frac{2 \sqrt{5}}{5}$. 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) From the problem, we know
$$
\begin{aligned}
|\vec{a}-\vec{b}|^{2} & =(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2} \\
& =2-2(\cos \alpha \cos \beta+\sin \alpha \sin \beta) \\
& =2-2 \cos (\alpha-\beta),
\end{aligned}
$$

then $2-2 \cos (\alpha-\beta)=\frac{4}{5}, \cos (\alpha-\beta)=\frac{3}{5}$. 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 $\cos \beta=\frac{12}{13}$, $\sin (\alpha-\beta)=\frac{4}{5}$, thus we get
$$
\begin{aligned}
\sin \alpha & =\sin [\beta+(\alpha-\beta)] \\
& =\sin \beta \cos (\alpha-\beta)+\cos \beta \sin (\alpha-\beta) \\
& =-\frac{5}{13} \times \frac{3}{5}+\frac{12}{13} \times \frac{4}{5} \\
& =\frac{33}{65} .
\end{aligned}
$$

Therefore, the answer is C.