11 Let $\{a_{n}\}$ be a sequence with the sum of the first $n$ terms denoted as $S_{n}$, and satisfying: $S_{2}=3, 2 S_{n}=n+n a_{n}$ $\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula for the sequence $\{a_{n}\}$;
(2) Find the sum of the first $2 n$ terms $T_{2 n}$ of the sequence $c_{n}=\left\{\begin{array}{ll}a_{n+1}, & n \text { is odd, } \\ 3 \times 2^{a_{n-1}}+1, & n \text { is even }\end{array}\right.$.
11 (1) Since $2 S_{n}=n+n a_{n}$, therefore, when $n=1$, from $2 S_{1}=1+a_{1}$, we get $a_{1}=1$;  
When $n \geqslant 2$, from $2 S_{n}=n+n a_{n}$, we get $2 S_{n-1}=n-1+(n-1) a_{n-1}$.  
Subtracting the two equations, we get  
$$
2 a_{n}=1+n a_{n}-(n-1) a_{n-1} \text {. }
$$

So
$$
2 a_{n+1}=1+(n+1) a_{n+1}-n a_{n} .
$$
(2) - (1), we get
$$
(n-1) a_{n+1}+(n-1) a_{n-1}=2(n-1) a_{n},
$$

which means when $n \geqslant 2$, $a_{n+1}+a_{n-1}=2 a_{n}$.  
Also, $S_{2}=3, a_{1}=1 \Rightarrow a_{2}=2$, so the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. Therefore, $a_{n}=n\left(n \in \mathbf{N}^{*}\right)$.
$$
\text { (2) } \begin{aligned}
T_{2 n} & =\left(a_{2}+a_{4}+\cdots+a_{2 n}\right)+3\left(2^{1}+2^{3}+\cdots+2^{2 n-1}\right)+n \\
& =n(n+1)+2^{2 n+1}-2+n \\
& =2^{2 n+1}+n^{2}+2 n-2 .
\end{aligned}
$$