若 $\alpha, \beta$ 均为锐角, $\sin \alpha=\frac{2 \sqrt{5}}{5}, \sin (\alpha+\beta)=\frac{3}{5}$, 则 $\cos \beta=$
$\because \alpha$ 为锐角, $\sin \alpha=\frac{2 \sqrt{5}}{5}>\frac{\sqrt{2}}{2} \mathrm{~s}, \therefore \alpha>45^{\circ}$ 且 $\cos \alpha=\frac{\sqrt{5}}{5}$,
   $\because \sin (\alpha+\beta)=\frac{3}{5}$, 且 $\frac{1}{2}<\frac{3}{5}<\frac{\sqrt{2}}{2}, \therefore \frac{\pi}{2}<\alpha+\beta<\pi$,

$\therefore \cos (\alpha+\beta)=-\frac{4}{5}$, 则 $\cos \beta=\cos [(\alpha+\beta)-\alpha]=\cos (\alpha+\beta) \cos \alpha+\sin (\alpha+\beta)$

$\sin \alpha=-\frac{4}{5} \times \frac{\sqrt{5}}{5}+\frac{3}{5} \times \frac{2 \sqrt{5}}{5}=\frac{2 \sqrt{5}}{25}$. 故选 B.