11. Given $f(x)=\frac{a x+1}{3 x-1}$, and the equation $f(x)=-4 x+8$ has two distinct positive roots, one of which is three times the other. Let the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be $S_{n}$ and $T_{n}$ respectively, and $\frac{S_{n}}{T_{n}}=f(n)(n=1,2, \cdots)$.
(1) If $g(n)=\frac{a_{n}}{b_{n}}(n=1,2, \cdots)$, find the maximum value of $g(n)$.
(2) If $a_{1}=\frac{5}{2}$, and the common difference of the sequence $\left\{b_{n}\right\}$ is 3, investigate whether there are equal terms in the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$. If so, find the general term formula of the sequence $\left\{c_{n}\right\}$ formed by these equal terms arranged in ascending order; if not, explain the reason.
(1) From $f(x)=-4 x+8$, we get $\frac{a x+1}{3 x-1}=-4 x+8$.
Rearranging gives
$$
12 x^{2}-(28-a) x+9=0 \text {. }
$$

Let $x_{1} 、 x_{2}$ be the two roots of equation (1), and $x_{2}=3 x_{1}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=4 x_{1}=\frac{28-a}{12}, \\
x_{1} x_{2}=3 x_{1}^{2}=\frac{9}{12} .
\end{array} .\right.
$$

Solving gives $x_{1}=\frac{1}{2}, a=4$.
Thus, $f(x)=\frac{4 x+1}{3 x-1}$.
Since $\frac{S_{n}}{T_{n}}=f(n)=\frac{4 n+1}{3 n-1}$, we have
$$
\begin{array}{l}
g(n)=\frac{a_{n}}{b_{n}}=\frac{S_{2 n-1}}{T_{2 n-1}}=\frac{4(2 n-1)+1}{3(2 n-1)-1} \\
=\frac{8 n-3}{6 n-4}=\frac{4}{3}+\frac{7}{3(6 n-4)} .
\end{array}
$$

Therefore, when $n=1$, $g(n)$ reaches its maximum value $\frac{5}{2}$.
(2) From (1), we know $\frac{a_{1}}{b_{1}}=\frac{5}{2}, \frac{a_{2}}{b_{2}}=\frac{13}{8}$.

Combining $a_{1}=\frac{5}{2}$ and the common difference of the sequence $\left\{b_{n}\right\}$ being 3, we get $b_{1}=1, b_{2}=4, a_{2}=\frac{13}{2}$.
Thus, $a_{n}=\frac{5}{2}+4(n-1)=4 n-\frac{3}{2}$,
$$
b_{n}=1+3(n-1)=3 n-2 \text {. }
$$

For any positive integer $n$, we know $a_{n} \notin \mathbf{Z}, b_{n} \in \mathbf{Z}$.
Therefore, there are no equal terms in the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$.