11. Given the sequence $\left\{x_{n}\right\}$ with the sum of the first $n$ terms $S_{n}=2 a_{n}-2\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula $a_{n}$ of $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n}}-\frac{1}{n(n+1)}, T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, find the positive integer $k$ such that for any $n \in \mathbf{N}^{*}$, $T_{k} \geqslant T_{n}$;
(3) Let $c_{n}=\frac{a_{n+1}}{\left(1+a_{n}\right)\left(1+a_{n+1}\right)}, R_{n}$ be the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$, if for any $n \in \mathbf{N}^{*}$, $R_{n}<\lambda$ holds, find the minimum value of $\lambda$.
(1) $S_{n}=2 a_{n}-2, S_{n+1}=2 a_{n+1}-2 \Rightarrow a_{n+1}=2\left(a_{n+1}-a_{n}\right) \Rightarrow a_{n+1}=2 a_{n}$, when $n=1$, $a_{1}=2 a_{1}-2 \Rightarrow a_{1}=2$. Therefore, $a_{n}=2^{n}\left(n \in \mathbf{N}^{*}\right)$.
(2) $b_{n}=\frac{1}{2^{n}}-\left(\frac{1}{n}-\frac{1}{n+1}\right) \Rightarrow T_{n}=1-\frac{1}{2^{n}}-\left(1-\frac{1}{n+1}\right)=\frac{1}{n+1}-\frac{1}{2^{n}}$.

Since $\left\{\begin{array}{l}\frac{1}{n+1}-\frac{1}{2^{n}} \geqslant \frac{1}{n}-\frac{1}{2^{n-1}}, \\ \frac{1}{n+1}-\frac{1}{2^{n}} \geqslant \frac{1}{n+2}-\frac{1}{2^{n+1}}\end{array} \Rightarrow\left\{\begin{array}{l}n(n+1) \geqslant 2^{n}, \\ (n+1)(n+2) \leqslant 2^{n+1}\end{array} \Rightarrow n=4\right.\right.$.
Therefore, there exists a unique positive integer $k=4$ such that for any $n \in \mathbf{N}^{*}$, $T_{k} \geqslant T_{n}$.
(3) $c_{n}=\frac{2^{n+1}}{\left(1+2^{n}\right)\left(1+2^{n+1}\right)}=2\left(\frac{1}{1+2^{n}}-\frac{1}{1+2^{n+1}}\right)$
$\Rightarrow R_{n}=2\left(\frac{1}{3}-\frac{1}{1+2^{n+1}}\right)<\lambda \Rightarrow \lambda \geqslant \frac{2}{3}$, so the minimum value of $\lambda$ is $\frac{2}{3}$.