8. 已知 $\cos \alpha=\frac{1}{3}$, 则 $\sin \left(\alpha-\frac{\pi}{2}\right) \cdot \cos \left(\frac{3 \pi}{2}+\alpha\right) \tan (\pi-\alpha)=$
$\sin \left(\alpha-\frac{\pi}{2}\right) \cos \left(\frac{3 \pi}{2}+\alpha\right) \tan (\pi-\alpha)$

$=-\cos \alpha \sin \alpha(-\tan \alpha)=\sin ^{2} \alpha=1-\cos ^{2} \alpha$

$=1-\left(\frac{1}{3}\right)^{2}=\frac{8}{9}$.

答案: $\frac{8}{9}$