10. Let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, and $a_{1}+a_{2}=4, \frac{2 S_{n+1}+1}{2 S_{n}+1}=\frac{a_{2}}{a_{1}}=c\left(c>0, n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n} \log _{3} a_{n}$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $T_{n}$.
(1) $\frac{2\left(a_{1}+a_{2}\right)+1}{2 a_{1}+1}=\frac{a_{2}}{a_{1}} \Rightarrow \frac{9}{2 a_{1}+1}=\frac{a_{2}}{a_{1}} \Rightarrow 9 a_{1}=\left(2 a_{1}+1\right)\left(4-a_{1}\right)$
$\Rightarrow a_{1}^{2}+a_{1}-2=0 \Rightarrow a_{1}=1$ or $a_{1}=-2$ (discard). Thus $a_{2}=3 \Rightarrow c=3$.
Therefore, $2 S_{n+1}+1=6 S_{n}+3 \Rightarrow S_{n+1}=3 S_{n}+1$
$\Rightarrow S_{n+1}+\frac{1}{2}=3\left(S_{n}+\frac{1}{2}\right) \Rightarrow S_{n}+\frac{1}{2}=\frac{1}{2} \cdot 3^{n} \Rightarrow S_{n}=\frac{3^{n}-1}{2}$, so $a_{n}=3^{n-1}\left(n \in \mathbf{N}^{*}\right)$.
(2) $b_{n}=(n-1) \cdot 3^{n-1}=[k(n+1)+b] \cdot 3^{n}-(k n+b) \cdot 3^{n-1}$,

then $n-1=2 k n+3 k+2 b \Rightarrow k=\frac{1}{2}, b=-\frac{5}{4}$.
So $T_{n}=\sum_{k=1}^{n} b_{k}=\left[\frac{1}{2}(n+1)-\frac{5}{4}\right] \cdot 3^{n}-\left(\frac{1}{2}-\frac{5}{4}\right)=\frac{(2 n-3) \cdot 3^{n}+3}{4}$.