11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms is
$$
S_{n}=2 a_{n}-2\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the general term formula $a_{n}$;
(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 $n \in$ $\mathbf{Z}_{+}$, we have $T_{k} \geqslant T_{n}$;
(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 $n \in \mathbf{Z}_{+}$, we have $R_{n}<\lambda$, find the minimum value of $\lambda$.
(1) From $S_{n}=2 a_{n}-2$, we get $S_{n+1}=2 a_{n+1}-2$.
Subtracting the two equations yields
$$
a_{n+1}=2 a_{n+1}-2 a_{n} \Rightarrow a_{n+1}=2 a_{n} \text {. }
$$

Thus, $\left\{a_{n}\right\}$ is a geometric sequence with common ratio $q=2$.
From $S_{1}=2 a_{1}-2 \Rightarrow a_{1}=2 a_{1}-2 \Rightarrow a_{1}=2$.
Therefore, $a_{n}=2^{n}$.
(2) From (1), we know
$$
\begin{aligned}
b_{n} & =\frac{1}{2^{n}}-\frac{1}{n(n+1)} \\
& =\frac{1}{n(n+1)}\left(\frac{n(n+1)}{2^{n}}-1\right) .
\end{aligned}
$$

Calculations show that $b_{1}=0, b_{2}>0, b_{3}>0, b_{4}>0$.
For $n \geqslant 5$, we have
$$
\begin{array}{l}
\frac{n(n+1)}{2^{n}}-\frac{(n+1)(n+2)}{2^{n+1}} \\
=\frac{(n+1)(n-2)}{2^{n+1}}>0,
\end{array}
$$

indicating that for $n \geqslant 5$, $\left\{\frac{n(n+1)}{2^{n}}\right\}$ is a decreasing sequence.
Thus, for $n \geqslant 5$,
$$
\frac{n(n+1)}{2^{n}} \leqslant \frac{5(5+1)}{2^{5}}T_{5}>\cdots$.
Therefore, for any $n \in \mathbf{Z}_{+}$, we have $T_{4} \geqslant T_{n}$. Hence, $k=4$.
(3) From (1), we know
$$
\begin{aligned}
c_{n} & =\frac{a_{n+1}}{\left(1+a_{n}\right)\left(1+a_{n+1}\right)} \\
& =\frac{2^{n+1}}{\left(1+2^{n}\right)\left(1+2^{n+1}\right)} \\
= & 2\left(\frac{1}{2^{n}+1}-\frac{1}{2^{n+1}+1}\right) \\
\Rightarrow & R_{n}=2 \sum_{k=1}^{n}\left(\frac{1}{2^{k}+1}-\frac{1}{2^{k+1}+1}\right) \\
& =2\left(\frac{1}{3}-\frac{1}{2^{n+1}+1}\right)=\frac{2}{3}-\frac{2}{2^{n+1}+1} .
\end{aligned}
$$

Since for any $n \in \mathbf{Z}_{+}$, we have $R_{n}<\lambda$, it follows that $\lambda \geqslant \frac{2}{3}$.

Thus, the minimum value of $\lambda$ is $\frac{2}{3}$.