215 * Given $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$. Then $\frac{1+\sin \alpha+\cos \alpha}{1+\sin \alpha-\cos \alpha}=$ $\qquad$
From $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, we have $\left(\tan \frac{\alpha}{2}-2\right)^{2}=1+\tan ^{2} \frac{\alpha}{2}$. Therefore, $\tan \frac{\alpha}{2}=\frac{3}{4}$. Also, $\tan \frac{a}{2}=\frac{1+\cos \alpha}{\sin \alpha}=\frac{\sin \alpha}{1-\cos \alpha}=\frac{1+\cos \alpha+\sin \alpha}{\sin \alpha+1-\cos \alpha}$, so the original expression $=\frac{3}{4}$. Furthermore, when $\alpha=(4 k+1) \pi$, $\frac{1+\sin \pi+\cos \pi}{1+\sin \pi-\cos \pi}=0$, so the original expression $=0$. Therefore, the original expression equals $\frac{3}{4}$ or 0.