7. 已知: $\cos (2 \alpha-\beta)=-\frac{\sqrt{2}}{2}, \sin (\alpha-2 \beta)=\frac{\sqrt{2}}{2}$, 且 $\frac{\pi}{4}<\alpha<\frac{\pi}{2}, \quad 0<\beta<\frac{\pi}{4}$, 求 $\cos (\alpha+\beta)$.
因为 $\frac{\pi}{4}<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{4}$, 所以 $\frac{\pi}{4}<2 \alpha-\beta<\pi$.

因为 $\cos (2 \alpha-\beta)=-\frac{\sqrt{2}}{2}$, 所以 $\frac{\pi}{2}<2 \alpha-\beta<\pi$.

所以 $\sin (2 \alpha-\beta)=\frac{\sqrt{2}}{2}$.

因为 $\frac{\pi}{4}<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{4}$, 所以 $-\frac{\pi}{4}<\alpha-2 \beta<\frac{\pi}{2}$,
因为 $\sin (\alpha-2 \beta)=\frac{\sqrt{2}}{2}$, 所以 $0<\alpha-2 \beta<\frac{\pi}{2}$,

所以 $\cos (\alpha-2 \beta)=\frac{\sqrt{2}}{2}$.

所以 $\cos (\alpha+\beta)=\cos [(2 \alpha-\beta)-(\alpha-2 \beta)]$

$=\cos (2 \alpha-\beta) \cos (\alpha-2 \beta)+\sin (2 \alpha-\beta) \cdot \sin (\alpha-2 \beta)$

$=-\frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}=0$.