15. For two functions $f(x)$ and $g(x)$ defined on the interval $[p, q]$, if for any $x \in [p, q]$, we have $|f(x) - g(x)| \leqslant 1$, then $f(x)$ and $g(x)$ are said to be close functions on the interval $[p, q]$. Otherwise, they are said to be non-close functions on $[p, q]$. Now, consider two functions $f_{1}(x) = \log _{a}(x - 3a)$ and $f_{2}(x) = \log _{a} \frac{1}{x - a}$ (where $a > 0$ and $a \neq 1$), and a given interval $[a+2, a+3]$.
(1) If $f_{1}(x)$ and $f_{2}(x)$ are both defined on $[a+2, a+3]$, find the range of values for $a$;
(2) Discuss whether $f_{1}(x)$ and $f_{2}(x)$ are close functions on the interval $[a+2, a+3]$.
15. (1) Clearly $a>0, a \neq 1$, and $\left\{\begin{array}{l}x-3 a>0, \\ x-a>0,\end{array}\right.$ then $x>3 a$, and $f_{1}(x) 、 f_{2}(x)$ are defined on $[a+2, a+3]$ if and only if $a+2>3 a$, thus $0<a<1$.
(2) $\left|f_{1}(x)-f_{2}(x)\right|=\left|\log _{a}\left(x^{2}-4 a x+3 a^{2}\right)\right|=\mid \log _{a}\left[(x-2 a)^{2}-a^{2}\right]$, when $x \in[a+2, a+3]$, $(x-2 a)^{2}-a^{2} \in[4-4 a, 9-6 a]$ Let $g(x)=\log _{a}\left(x^{2}-4 a x+3 a^{2}\right)$, then $g(x)_{\text {min }}=g(a+3)=$ $\log _{a}(9-6 a), g(x)_{\max }=g(a+2)=\log _{a}(4-4 a)$.
To make $\left|f_{1}(x)-f_{2}(x)\right| \leqslant 1$, it must be that $\left\{\begin{array}{l}0<a<1, \\ \log _{a}(9-6 a) \geqslant-1 \\ \log _{a}(4-4 a) \leqslant 1\end{array}\right.$
Solving this, we get $a \in\left(0, \frac{9-\sqrt{57}}{12}\right]$, that is, when $a \in\left(0, \frac{9-\sqrt{57}}{12}\right]$, $f_{1}(x)$ and $f_{2}(x)$ are close; when $a \in\left(\frac{9-\sqrt{57}}{12}, 1\right)$, $f_{1}(x)$ and $f_{2}(x)$ are not close.