14. For two functions $f(x)$ and $g(x)$ defined on the interval $[m, n]$, if for any $x \in [m, n]$ we have $|f(x)-g(x)| \leqslant 1$, then $f(x)$ and $g(x)$ are said to be close on $[m, n]$, otherwise $f(x)$ and $g(x)$ are said to be not close on $[m, n]$. Now consider two functions $f_{1}(x)=\log _{a}(x-3 a)$ and $f_{2}(x)=\log _{a} \frac{1}{x-a}(a>0, a \neq 1)$, given the interval $[a+2, a+3]$.
(1) If $f_{1}(x)$ and $f_{2}(x)$ are both defined on the given interval $[a+2, a+3]$, find the range of values for $a$.
(2) Discuss whether $f_{1}(x)$ and $f_{2}(x)$ are close on the given interval $[a+2, a+3]$.
14. (1) $\left\{\begin{array}{l}x-3 a>0 \\ \frac{1}{x-a}>0,\end{array}\right.$ we get $\left\{\begin{array}{l}x>3 a \\ x>a .\end{array}\right.$ Since $a>0$, then $x>3 a$. Also, $f_{1}(x)$ and $f_{2}(x)$ are defined on the interval $[a+2, a+3]$, so $\left\{\begin{array}{l}a+2>3 a \\ a>0,\end{array}\right.$ which gives $0<a<1$.
(2) If $\left|f_{1}(x)-f_{2}(x)\right|=\left|\log _{a}(x-3 a)(x-a)\right| \leqslant 1$, then $-1 \leqslant \log _{a}(x-3 a)(x-a) \leqslant$ 1. Since $0<a<1$, we have $\left\{\begin{array}{l}(x-3 a)(x-a) \geqslant a \\ (x-3 a)(x-a) \leqslant \frac{1}{a},\end{array}\right.$ which simplifies to $\left\{\begin{array}{l}x^{2}-4 a x+3 a^{2}-a \geqslant 0 \\ x^{2}-4 a x+3 a^{2}-\frac{1}{a} \leqslant 0 .\end{array}\right.$ The axis of symmetry is $x=2 a<a+2$, so $\left\{\begin{array}{l}(a+2)^{2}-4 a(a+2)+3 a^{2}-a \geqslant 0 \\ (a+2)^{2}-4 a(a+2)+3 a^{2}-\frac{1}{a} \leqslant 0,\end{array}\right.$ solving this gives $0<a \leqslant \frac{9-\sqrt{57}}{12}$. Therefore, when $0<a \leqslant \frac{9-\sqrt{57}}{12}$, $f_{1}(x)$ and $f_{2}(x)$ are close on the interval $[a+2, a+3]$, otherwise they are not close.