2. If $a$ is a positive real number, and
$$
f(x)=\log _{2}\left(a x+\sqrt{2 x^{2}+1}\right)
$$

is an odd function, then the solution set for $f(x)>\frac{3}{2}$ is $\qquad$ .
2. $\left(\frac{7}{8},+\infty\right)$.

Given $f(x)+f(-x)=0$
$$
\begin{array}{l}
\Rightarrow\left(a x+\sqrt{2 x^{2}+1}\right)\left(-a x+\sqrt{2 x^{2}+1}\right)=1 \\
\Rightarrow\left(2-a^{2}\right) x^{2}=0 \Rightarrow a=\sqrt{2} .
\end{array}
$$

Since $\sqrt{2} x+\sqrt{2 x^{2}+1}$ is an increasing function on the interval $(0,+\infty)$, thus, $f(x)$ is an increasing function on the interval $(0,+\infty)$. 
Combining the fact that $f(x)$ is an odd function, we know that $f(x)$ is an increasing function on $\mathbf{R}$.
Then the equation $f(x)=\frac{3}{2}$
$$
\begin{array}{l}
\Leftrightarrow \sqrt{2} x+\sqrt{2 x^{2}+1}=2^{\frac{3}{2}}=2 \sqrt{2} \\
\Leftrightarrow \sqrt{2 x^{2}+1}=\sqrt{2}(2-x) \\
\Rightarrow x=\frac{7}{8} .
\end{array}
$$

Therefore, the solution set for $f(x)>\frac{3}{2}$ is $\left(\frac{7}{8},+\infty\right)$.