1. Given that the circumradius of $\triangle A B C$ is $R$, and
$$
2 R\left(\sin ^{2} A-\sin ^{2} C\right)=(\sqrt{2} a-b) \sin B,
$$

where $a$ and $b$ are the sides opposite to $\angle A$ and $\angle B$ respectively. Then the size of $\angle C$ is $\qquad$.
$-1.45^{\circ}$.
From the given equation and using the Law of Sines, we get
$$
\begin{array}{l}
a^{2}-c^{2}=(\sqrt{2} a-b) b \\
\Rightarrow a^{2}+b^{2}-c^{2}=\sqrt{2} a b \\
\Rightarrow \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{\sqrt{2}}{2} \\
\Rightarrow \angle C=45^{\circ} .
\end{array}
$$