1. Given that $f(x)$ is a function defined on $\mathbf{R}$, and for any real number $x$, we have $2 f(x)+f\left(x^{2}-1\right)=1$, then the value of $f(\sqrt{2})$ is $\qquad$ .
Notice that $2 f(-x)+f\left(x^{2}-1\right)=1 \Rightarrow f(-x)=f(x)$, then we have $\left\{\begin{array}{l}2 f(\sqrt{2})+f(1)=1, \\ 2 f(1)+f(0)=1, \\ 2 f(0)+f(-1)=1\end{array} \Rightarrow\left\{\begin{array}{l}f(\sqrt{2})=\frac{1}{3}, \\ f(1)=f(0)=\frac{1}{3} .\end{array}\right.\right.$