11. Insert the sum of each pair of adjacent terms between them in a sequence, forming a new sequence, which is called a "Z-extension" of the sequence. Given the sequence $1,2,3$, after the first "Z-extension" it becomes $1,3,2,5,3$; after the second "Z-extension" it becomes $1,4,3,5,2,7,5,8,3 ; \cdots$; after the $n$-th "Z-extension" it becomes $1, x_{1}, x_{2}, \cdots, x_{m}, 3$, and let $a_{n}=1+x_{1}+x_{2}+\cdots+x_{m}+3$.
(1) Find the values of $a_{1}, a_{2}, a_{3}$;
(2) If $b_{n}=a_{n}-2$, prove that the sequence $\left\{b_{n}\right\}$ is a geometric sequence, and find the general term formula for the sequence $\left\{a_{n}\right\}$.
$$
\begin{array}{l}
\text { (1) } a_{1}=1+3+2+5+3=14, a_{2}=1+4+3+5+2+7+5+8+3=38 \\
a_{3}=1+5+4+7+3+8+5+7+2+9+7+12+5+13+8+11+3=110
\end{array}
$$
(2) From (1), we know that $b_{1}=12, b_{2}=36, b_{3}=108$, conjecture $b_{n}=4 \cdot 3^{n}\left(n \in \mathbf{N}^{*}\right)$.
When $n=1,2$, the conclusion holds. Assume that when $n=k$, the conclusion holds, i.e.,
$$
a_{k}=1+x_{1}+x_{2}+\cdots+x_{p}+3=4 \cdot 3^{k}+2 \text {, }
$$

then $a_{k+1}=1+\left(1+x_{1}\right)+x_{1}+\left(x_{1}+x_{2}\right)+x_{2}+\cdots+x_{p}+\left(x_{p}+3\right)+3$ $=8+3\left(x_{1}+x_{2}+\cdots+x_{p}\right)=-4+3 a_{k}=4 \cdot 3^{k+1}+2$, so when $n=k+1$, the conclusion also holds. In summary, we have $b_{n}=4 \cdot 3^{n}, a_{n}=4 \cdot 3^{n}+2\left(n \in \mathbf{N}^{*}\right)$.