已知, $\cos \alpha=\frac{\sqrt{5}}{5}, \sin (\alpha-\beta)=\frac{\sqrt{10}}{10}$, 且 $\alpha 、 \beta \in\left(0, \frac{\pi}{2}\right)$, 求:

(1) $\cos (2 \alpha-\beta)$ 的值;

(2) $\beta$ 的值.
(1) 因为 $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, 所以 $\alpha-\beta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,

又因为 $\sin (\alpha-\beta)=\frac{\sqrt{10}}{10}$, 则 $\cos (\alpha-\beta)=\frac{3 \sqrt{10}}{10}$, 而 $\sin \alpha=\frac{2 \sqrt{5}}{5}$,

$\cos (2 \alpha-\beta)=\cos [\alpha+(\alpha-\beta)]=\cos \alpha \cos (\alpha-\beta)-\sin \alpha \sin (\alpha-\beta)=\frac{\sqrt{2}}{10}$,

(2) $\cos \beta=\cos [\alpha-(\alpha-\beta)]=\cos \alpha \cos (\alpha-\beta)+\sin \alpha \sin (\alpha-\beta)=\frac{\sqrt{2}}{2}$,

又 $\because \beta \in\left(0, \frac{\pi}{2}\right), \therefore \beta=\frac{\pi}{4}$.