13. Let $\{a_{n}\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}, S_{n}=2^{n}+r$ (where $r$ is a constant), and let $b_{n}=2\left(1+\log _{2} a_{n}\right)\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the sum of the first $n$ terms of the sequence $\{a_{n} b_{n}\}$, denoted as $T_{n}$;
(2) If for any positive integer $n$, $\frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdots \frac{1+b_{n}}{b_{n}} \geqslant k \sqrt{n+1}$ holds, find the maximum value of the real number $k$.
(1) When $n \geqslant 2$, $a_{n}=S_{n}-S_{n-1}=2^{n-1}$, then $n=1 \Rightarrow r=-1$. Also, $b_{n}=2 n$,

thus $a_{n} b_{n}=n \cdot 2^{n}=[(n+1)-2] \cdot 2^{n+1}-(n-2) \cdot 2^{n}$,
so $T_{n}=\sum_{k=1}^{n} a_{k} b_{k}=[(n+1)-2] \cdot 2^{n+1}+2=(n-1) \cdot 2^{n+1}+2$.
(2) Let $G(n)=\frac{\left(1+b_{1}\right)\left(1+b_{2}\right) \cdots\left(1+b_{n}\right)}{b_{1} b_{2} \cdots b_{n} \sqrt{n+1}} \Rightarrow \frac{G(n+1)}{G(n)}=\frac{\left(1+b_{n+1}\right) \sqrt{n+1}}{b_{n+1} \sqrt{n+2}}$
$=\frac{(2 n+3) \sqrt{n+1}}{(2 n+2) \sqrt{n+2}} \geqslant 1 \Rightarrow n \geqslant-1$, then $G(n)$ is monotonically increasing, so $G(n)_{\min }=G(1)=\frac{3}{2 \sqrt{2}}=\frac{3 \sqrt{2}}{4}$. Thus, the maximum value of $k$ is $\frac{3 \sqrt{2}}{4}$.