11 Given $f(x)=\frac{a x+1}{3 x-1}$, the equation $f(x)=-4 x+8$ has two distinct positive roots, and one root is three times the other. The first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ have sums $S_{n}$ and $T_{n}$, respectively, and $\frac{S_{n}}{T_{n}}=f(n)(n=1,2,3, \cdots)$.
(1) Let $g(n)=\frac{a_{n}}{b_{n}}(n=1,2,3, \cdots)$, 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 there are, 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.
11 (1) From $f(x)=-4 x+8$, we get $\frac{a x+1}{3 x-1}=-4 x+8$, rearranging gives
$$
12 x^{2}-(28-a) x+9=0 .
$$

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

Since $x_{1}, x_{2}>0$, we have $x_{1}=\frac{1}{2}, a=4, f(x)=\frac{4 x+1}{3 x-1}$. Because
$$
\frac{S_{n}}{T_{n}}=f(n)=\frac{4 n+1}{3 n-1},
$$

we have
$$
\begin{aligned}
g(n) & =\frac{a_{n}}{b_{n}}=\frac{S_{2 n-1}}{T_{2 n-1}}=\frac{4(2 n-1)+1}{3(2 n-1)-1} \\
& =\frac{8 n-3}{6 n-4}=\frac{4}{3}+\frac{7}{3(6 n-4)} .
\end{aligned}
$$

Therefore, when $n=1$, $g(n)$ reaches its maximum value $\frac{5}{2}$.
(2) From (1), we know $\frac{a_{1}}{b_{1}}=\frac{5}{2}$, and given $a_{1}=\frac{5}{2}$, we have $b_{1}=1$. Since the common difference of the sequence $\left\{b_{n}\right\}$ is 3, we have
$$
b_{n}=1+3(n-1)=3 n-2 \text {. }
$$

From $g(n)=\frac{8 n-3}{6 n-4}$, we get
$$
a_{n}=\frac{5}{2}+4(n-1)=\frac{8 n-3}{2} .
$$

If there are equal terms in the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, let $a_{m}=b_{k}$ (where $m, k$ are positive integers), then $\frac{8 m-3}{2}=3 k-2$, rearranging gives $6 k-8 m=1$. Since $6 k-8 m$ is even, and 1 is odd, the above equation has no positive integer solutions.
Therefore, there are no equal terms in the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$.