1. In $\triangle A C D$, $\angle A=45^{\circ}, C D=7, B$ is a point on $A D$, $C B=5, B D=$ $3, A C=$ $\qquad$ .
1. $\frac{5 \sqrt{6}}{2}$.

In $\triangle C B D$, $\cos \angle C B D=\frac{B C^{2}+B D^{2}-C D^{2}}{2 B C \cdot B D}=-\frac{1}{2}$, so $\angle C B D=120^{\circ} \angle C B A=$ $120^{\circ}$. In $\triangle A B C$, $A C=\frac{B C \sin \angle C B A}{\sin A}=\frac{5 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}=\frac{5 \sqrt{6}}{2}$.