11. (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{2}=3, \\
2 a_{n+1}=3 a_{n}-a_{n-1}(n \geqslant 2) .
\end{array}
$$

Find (1) the general term formula $a_{n}$ of the sequence $\left\{a_{n}\right\}$;
(2) all positive integer values of $m$ that satisfy the inequality $\frac{a_{n}-m}{a_{n+1}-m}<\frac{2}{3}$.
11. (1) From $2 a_{n+1}=3 a_{n}-a_{n-1}$, we know
$$
2\left(a_{n+1}-a_{n}\right)=a_{n}-a_{n-1} \text {. }
$$

Thus, the sequence $\left\{a_{n}-a_{n-1}\right\}$ is a geometric sequence with the first term $a_{2}-a_{1}=1$ and the common ratio $\frac{1}{2}$.
Therefore, $a_{n}-a_{n-1}=\left(\frac{1}{2}\right)^{n-2}$.
Summing up, we get $a_{n}=4-\left(\frac{1}{2}\right)^{n-2}$.
(2) From (1), the required inequality is
$$
\frac{4-\left(\frac{1}{2}\right)^{n-2}-m}{4-\left(\frac{1}{2}\right)^{n-1}-m}<\frac{2}{3} \text {. }
$$

Clearly, $m \geqslant 4$ has no solution.
Thus, it is easy to find $(m, n)=(1,1),(2,1),(3,2)$.