11. (20 points) Let $\{a_n\}$ be an arithmetic sequence with a non-zero common difference, satisfying $a_3 + a_6 = a_9$ and $a_5 + a_7^2 = 6a_9$. Let the sequence $\{b_n\}$ have the sum of its first $n$ terms denoted by $S_n$, and $4S_n + 2b_n = 3$. For any $i \in \mathbf{Z}_+$, insert $i$ numbers $x_{i1}, x_{i2}, \cdots, x_{ii}$ between $b_i$ and $b_{i+1}$, such that $b_i, x_{i1}, x_{i2}, \cdots, x_{ii}, b_{i+1}$ form an arithmetic sequence. Let $T_n = \sum_{i=1}^{n} \sum_{j=1}^{i} x_{ij}$. Does there exist positive integers $m$ and $n$ such that $T_n = \frac{a_{m+1}}{2a_m}$? If so, find all such pairs of positive integers $(m, n)$; if not, explain why.
11. Let the common difference of the sequence $\left\{a_{n}\right\}$ be $d(d \neq 0)$.

Then from the condition $a_{3}+a_{6}=a_{9}$, we get
$$
\begin{array}{l}
\left(a_{1}+2 d\right)+\left(a_{1}+5 d\right)=a_{1}+8 d \\
\Rightarrow a_{1}=d .
\end{array}
$$

Also, from $a_{5}+a_{7}^{2}=6 a_{9}$, we get
$$
\left(a_{1}+4 d\right)+\left(a_{1}+6 d\right)^{2}=6\left(a_{1}+8 d\right) \text {. }
$$

Substituting $a_{1}=d$ into the above equation, we get $49 d^{2}=49 d$.
Since $d \neq 0$, then
$$
d=1 \Rightarrow a_{n}=n \text {. }
$$
$$
\begin{array}{l}
\text { From } 4 S_{n}+2 b_{n}=3, \text { for } n \geqslant 2 \text {, } \\
4 S_{n-1}+2 b_{n-1}=3 .
\end{array}
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
4 b_{n}+2 b_{n}-2 b_{n-1}=0 \\
\Rightarrow b_{n}=\frac{1}{3} b_{n-1}(n \geqslant 2) . \\
\text { Also, } 4 b_{1}+2 b_{1}=3 \text {, then } b_{1}=\frac{1}{2} .
\end{array}
$$

Thus, $\left\{b_{n}\right\}$ is a geometric sequence with the first term $\frac{1}{2}$ and common ratio $\frac{1}{3}$. Therefore, $b_{n}=\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}\left(n \in \mathbf{Z}_{+}\right)$.

Inserting $x_{n 1}, x_{n 2}, \cdots, x_{n n}$ between $b_{n}$ and $b_{n+1}$, since $b_{n}, x_{n 1}, \cdots, x_{n n}, b_{n+1}$ form an arithmetic sequence, let the common difference be $d$. Hence,
$$
\begin{aligned}
d_{n} & =\frac{b_{n+1}-b_{n}}{(n+2)-1} \\
= & \frac{\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{n}-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{n-1}}{n+1} \\
= & -\frac{1}{3^{n}(n+1)} \\
\Rightarrow & x_{n k}=b_{n}+k d_{n}=\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}-\frac{k}{3^{n}(n+1)} \\
\Rightarrow & \sum_{k=1}^{n} x_{n k} \\
& =\frac{1}{2}\left(\frac{1}{3}\right)^{n-1} n-\frac{1}{3^{n}(n+1)} \cdot \frac{n(n+1)}{2} \\
& =\frac{n}{3^{n}} \\
\Rightarrow & T_{n}=\sum_{i=1}^{n} \sum_{j=1}^{i} x_{i j}=\sum_{k=1}^{n} \frac{k}{3^{k}} \\
\Rightarrow & \frac{1}{3} T_{n}=\sum_{k=1}^{n} \frac{k}{3^{k+1}} .
\end{aligned}
$$

Subtracting the above two equations, we get
$$
\begin{array}{l}
\frac{2}{3} T_{n}=\sum_{k=1}^{n} \frac{1}{3^{k}}-\frac{n}{3^{n+1}}=\frac{1}{2}\left(1-\frac{1}{3^{n}}\right)-\frac{n}{3^{n+1}} \\
\Rightarrow T_{n}=\frac{3}{4}-\frac{1}{4 \times 3^{n-1}}-\frac{n}{2 \times 3^{n-1}} .
\end{array}
$$

If $T_{n}=\frac{a_{m+1}}{2 a_{m}}$, since $a_{n}=n$, then $a_{m}=m$.
Thus, $\frac{3}{4}-\frac{1}{4 \times 3^{n-1}}-\frac{n}{2 \times 3^{n}}$
$$
=\frac{m+1}{2 m}=\frac{1}{2}+\frac{1}{2 m} \text {. }
$$

Therefore, $m=\frac{2 \times 3^{n}}{3^{n}-2 n-3}$
$$
\begin{array}{l}
=\frac{2\left(3^{n}-2 n-3\right)+4 n+6}{3^{n}-2 n+6} \\
=2+\frac{4 n+6}{3^{n}-2 n-6} .
\end{array}
$$

When $n=1$, $m=2+\frac{10}{-2}=-3 \notin \mathbf{Z}_{+}$;
When $n=2$, $m=2+\frac{14}{2}=9 \in \mathbf{Z}_{+}$;
When $n=3$, $m=2+1=3 \in \mathbf{Z}_{+}$.
We now prove: For $n \geqslant 4\left(n \in \mathbf{z}_{+}\right)$,
$$
3^{n}-2 n-3>4 n+6,
$$

which is equivalent to proving $3^{n}-6 n-9>0$.
$$
\begin{array}{l}
\text { Let } f(x)=3^{x}-6 x-9(x \geqslant 4) \text {. Then } \\
f^{\prime}(x)=3^{x} \ln 3-6>3^{x}-6 \\
\geqslant 3^{4}-6=75>0 .
\end{array}
$$

Thus, $f(x)$ is monotonically increasing on the interval $[4,+\infty)$.
Therefore, for $n \geqslant 4$,
$$
3^{n}-6 n-9>3^{4}-6 \times 4-9=48>0,
$$

which implies $0<\frac{4 n+6}{3^{n}-2 n-3}<1$.
Thus, for $n \geqslant 4$, $m$ is not an integer.
Therefore, the integer pairs are $(9,2),(3,3)$.