1. Given $\triangle A B C$ with the three interior angles $\angle A, \angle B, \angle C$ opposite to the sides $a, b, c$ respectively, and satisfying $a \sin A \sin B + b \cos ^{2} A = \sqrt{2} a$. Then the value of $\frac{b}{a}$ is $\qquad$.
1. $\sqrt{2}$.

From the given and the Law of Sines, we get
$$
\sqrt{2} a=a \sin A \sin B+b\left(1-\sin ^{2} A\right)=b+\sin A(a \sin B-b \sin A)=b .
$$

Therefore, $\frac{b}{a}=\sqrt{2}$.