Four. (20 points) Insert the sum of each pair of adjacent terms between them in a sequence, forming a new sequence. Such an operation is called a “$Z$ expansion” of the sequence. Given the sequence $1,2,3$, the first $Z$ expansion results in the sequence $1,3,2,5,3$; the second $Z$ expansion results in the sequence $1,4,3,5,2,7,5,8,3$; $\cdots \cdots$ Let the sequence obtained after the $n$-th $Z$ expansion be $1, x_{1}, x_{2}$, $\cdots, x_{m}, 3$, and denote $a_{n}=1+x_{1}+x_{2}+\cdots+x_{m}+3$.
(1) Find the values of $a_{1} 、 a_{2} 、 a_{3}$;
(2) If $b_{n}=a_{n}-2$, prove that $\left\{b_{n}\right\}$ is a geometric sequence, and find the general term formula for the sequence $\left\{a_{n}\right\}$.
Let $x_{0}=1, x_{m+1}=3, a_{n}=\sum_{i=0}^{m+1} x_{i}$.
$$
\begin{array}{l}
\text { Then } a_{n+1}=\sum_{i=0}^{m+1} x_{i}+\sum_{i=0}^{m}\left(x_{i}+x_{i+1}\right) \\
=3 \sum_{i=0}^{m+1} x_{i}-x_{0}-x_{m+1} \\
=3 a_{n}-4 .
\end{array}
$$

Given $a_{1}=1+3+2+5+3=14$, we have
$$
a_{2}=3 a_{1}-4=38, a_{3}=3 a_{2}-4=110 \text {. }
$$

Thus, $b_{n+1}=a_{n+1}-2=3 a_{n}-6=3 b_{n}$. Also, $b_{1}=12 \neq 0$, so $\left\{b_{n}\right\}$ is a geometric sequence, and
$$
b_{n}=4 \times 3^{n} \text {. }
$$

Therefore, $a_{n}=4 \times 3^{n}+2$.