8. 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 of the inequality $f(x)>\frac{0}{2}$ is $\qquad$
Answer: $\left(\frac{7}{8},+\infty\right)$.
Analysis: From $f(x)+f(-x)=0$, we get
$$
\log _{2}\left(a x+\sqrt{2 x^{2}+1}\right)+\log _{2}\left(-a x+\sqrt{2 x^{2}+1}\right)=0
$$

which means
$$
\left(a x+\sqrt{2 x^{2}+1}\right)\left(-a x+\sqrt{2 x^{2}+1}\right)=1
$$

which simplifies to $\left(2-a^{2}\right) x^{2}=0$, so $a=\sqrt{2}$.
Since $\sqrt{2} x+\sqrt{2 x^{2}+1}$ is an increasing function on $(0,+\infty)$, $f(x)$ is also an increasing function on $(0,+\infty)$. Given that $f(x)$ is an odd function, $f(x)$ is an increasing function on $\mathbf{R}$. To solve the inequality $f(x)>\frac{3}{2}$, we only need to find the solution to the equation $f(x)=\frac{3}{2}$.
The equation $f(x)=\frac{3}{2}$ is equivalent to $\sqrt{2} x+\sqrt{2 x^{2}+1}=2^{\frac{3}{2}}=2 \sqrt{2}$, which simplifies to $\sqrt{2 x^{2}+1}=\sqrt{2}(2-x)$. Squaring both sides, we solve to get $x=\frac{7}{8}$. Therefore, the solution set of the inequality $f(x)>\frac{3}{2}$ is $\left(\frac{7}{8},+\infty\right)$.