13. Let $\{a_{n}\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and $S_{n}=2^{n}+r$ (where $r$ is a constant). Let $b_{n}=2\left(1+\log _{2} a_{n}\right) \quad\left(n \in \mathbf{N}^{*}\right)$.
(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 $n$, $\frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdots \cdot \frac{1+b_{n}}{b_{n}} \geq k \sqrt{n+1}$ holds,
Solution: (1) From the conditions, it is easy to know that $a_{1}=2+r, a_{2}=S_{2}-S_{1}=2, a_{3}=S_{3}-S_{2}=4$,
and from $a_{2}^{2}=a_{1} a_{3}$ we get $r=-1$.
Thus, $S_{n}=2^{n}-1$. Therefore, $a_{n}=2^{n-1}, b_{n}=2\left(1+\log _{2} a_{n}\right)=2 n, a_{n} b_{n}=n \cdot 2^{n}$.
Hence, $T_{n}=1 \times 2^{1}+2 \times 2^{2}+\cdots+(n-1) \times 2^{n-1}+n \times 2^{n}$
$$
2 T_{n}=1 \times 2^{2}+2 \times 2^{3}+\cdots+(n-1) \times 2^{n}+n \times 2^{n+1}
$$

From (1)-(2) we get: $-T_{n}=2^{1}+2^{2}+\cdots+2^{n-1}+2^{n}-n \times 2^{n+1}$, so $T_{n}=(n-1) \cdot 2^{n+1}+2$.
Therefore, the sum of the first $n$ terms of the sequence $\left\{a_{n} b_{n}\right\}$ is $T_{n}=(n-1) \cdot 2^{n+1}+2\left(n \in \mathbf{N}^{*}\right)$.
(2) Since $k \leq \frac{1}{\sqrt{n+1}} \cdot \frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdots \cdots \frac{1+b_{n}}{b_{n}}=\frac{1}{\sqrt{n+1}} \cdot \frac{1+2}{2} \cdot \frac{1+4}{4} \cdots \cdots \frac{1+2 n}{2 n}$,

Construct $f(n)=\frac{1}{\sqrt{n+1}} \cdot \frac{1+2}{2} \cdot \frac{1+4}{4} \cdots \cdots \frac{1+2 n}{2 n}$,
then $\frac{f(n+1)}{f(n)}=\frac{\sqrt{n+1}}{\sqrt{n+2}} \cdot \frac{1+2(n+1)}{2(n+1)}=\sqrt{\frac{4 n^{2}+12 n+9}{4 n^{2}+12 n+8}}>1$,
thus $\{f(n)\}$ is strictly increasing, so the minimum value of $f(n)$ is $f(1)=\frac{3}{4} \sqrt{2}$,
which means the maximum value of the real number $k$ is $\frac{3}{4} \sqrt{2}$.