1. Given in Rt $\triangle A B C$, $\angle C=90^{\circ}, \frac{A C}{B C}=$ $1+\frac{\sqrt{6}}{2}$. Then $\frac{A B}{A C}=$ .. $\qquad$
$$
1.2 \sqrt{2}-\sqrt{3} \text{. }
$$

Given $\tan B=\frac{A C}{B C}=1+\frac{\sqrt{6}}{2}$, then
$$
\frac{A B}{A C}=\frac{1}{\sin B}=\sqrt{1+\cot ^{2} B}=2 \sqrt{2}-\sqrt{3} \text{. }
$$