15. $x+\sqrt{x^{2}-2}>0$ implies $x>\sqrt{2}$, thus $f(x) \geqslant \sqrt{2}$. From the given, $x=\frac{a^{y}+2 a^{-y}}{2}$, which means $f^{-1}(x)=$ $\frac{a^{x}+2 a^{-x}}{2}$. When $a>1$, $x \in\left[\log _{a} \sqrt{2},+\infty\right)$; when $0<a<1$, $x \in\left(-\infty, \log _{a} \sqrt{2}\right]$. For $a>1$, $g(n)=\frac{\sqrt{2}}{2} f^{-1}\left(\log _{a} \sqrt{2}+n\right)=\frac{a^{n}+a^{-n}}{2}$. From $g(n)<$ $\frac{3^{n}+3^{-n}}{2}$, we get $a^{n}+a^{-n}<3^{n}+3^{-n}$, thus $1<a<3$.
15. Given the function $f(x)=\log _{a}\left(x+\sqrt{x^{2}-2}\right)(a>0, a \neq 1)$ with its inverse function $f^{-1}(x)$, let $g(n)=\frac{\sqrt{2}}{2} f^{-1}\left(n+\log _{a} \sqrt{2}\right)$. If $g(n)<\frac{3^{n}+3^{-n}}{2}(n \in$ $\left.\mathbf{N}_{+}\right)$, find the range of values for $a$.