1. If the function $f(x)=\log _{a}\left(x+\sqrt{b x^{2}+2 a^{2}}\right) \quad(a>0$ and $a \neq-1)$ is an odd function, then the real number pair $(a, b)=$ $\qquad$ .
1. $\left(\frac{\sqrt{2}}{2}, 1\right)$.

By the property of odd functions, we know $f(0)=\log _{a} \sqrt{2 a^{2}}=0$, which means $2 a^{2}=1$. Solving this, we get $a=\frac{\sqrt{2}}{2}$ (discarding the negative value), thus
$$
\begin{array}{l}
f(x)=\log _{\frac{\sqrt{2}}{2}}\left(x+\sqrt{b x^{2}+1}\right) \text {. Also, } \\
0=f(x)+f(-x)=\log _{\frac{\sqrt{2}}{2}}\left(x+\sqrt{b x^{2}+1}\right)+\log _{\frac{\sqrt{2}}{2}}\left(-x+\sqrt{b x^{2}+1}\right)=\log _{\frac{\sqrt{2}}{2}}\left((b-1) x^{2}+1\right),
\end{array}
$$

Thus, $(b-1) x^{2}=0$ always holds, hence $b=1$.