8. (1999 Henan Province Competition Question) Given $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, then $\frac{1+\sin \alpha+\cos \alpha}{1+\sin \alpha-\cos \alpha}=$
8. Fill in $\frac{3}{4}$ or 0. Reason: From $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, we have $\tan \frac{\alpha}{2}-2=\frac{1}{\cos \frac{\alpha}{2}}=\sec \frac{\alpha}{2}=\sqrt{1+\tan ^{2} \frac{\alpha}{2}}$, thus $\tan \frac{\alpha}{2}=\frac{3}{4}$. Also, $\tan \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=\frac{\sin \alpha}{1-\cos \alpha}=\frac{1+\cos \alpha+\sin \alpha}{\sin \alpha+1-\cos \alpha}$, hence the original expression $=\frac{3}{4}$.

Furthermore, when $\alpha=(4 k+1) \pi(k \in \mathbf{Z})$, it satisfies $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, thus, when $\alpha=(4 k+1) \pi$, $\frac{1+\cos \alpha+\sin \alpha}{1+\sin \alpha-\cos \alpha}=$ $\frac{1+\sin \pi+\cos \pi}{1+\sin \pi-\cos \pi}=0$.