1. Given $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$. Then
$$
\frac{1+\sin \alpha+\cos \alpha}{1+\sin \alpha-\cos \alpha}=
$$
$\qquad$
II. $1 . \frac{3}{4}$ or 0.
From $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, we have
$\operatorname{tg} \frac{\alpha}{2}-2=\sqrt{1+\operatorname{tg}^{2} \frac{\alpha}{2}}$.
Thus, $\operatorname{tg} \frac{a}{2}=\frac{3}{4}$.
Also, $\operatorname{tg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=\frac{\sin \alpha}{1-\cos \alpha}$
$$
=\frac{1+\cos \alpha+\sin \alpha}{\sin \alpha+1-\pi},
$$

Therefore, the original expression $=\frac{3}{4}$.
When $\alpha=(+k+1) \pi$, $\frac{1+\sin \pi+\cos \pi}{1+\sin \pi-\cos \pi}=0$, hence the original expression $=0$.