13. Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms $S_{n}=2^{n}+r$ (where $r$ is a constant). Let $b_{n}=2\left(1+\log _{2} a_{n}\right)\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the sum of the first $n$ terms $T_{n}$ of the sequence $\left\{a_{n} b_{n}\right\}$;
(2) If for any positive integer $n$, we have
$$
\frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdots \cdots \frac{1+b_{n}}{b_{n}} \geqslant k \sqrt{n+1} \text {, }
$$

find the maximum value of the real number $k$.
Three, 13. (1) From the conditions, we easily know
$$
\begin{array}{l}
a_{1}=2+r, a_{2}=S_{2}-S_{1}=2, \\
a_{3}=S_{3}-S_{2}=4 .
\end{array}
$$

Substituting into $a_{2}^{2}=a_{1} a_{3}$, we get $r=-1$.
Thus, $S_{n}=2^{n}-1$.
Then $a_{n}=2^{n-1}, b_{n}=2\left(1+\log _{2} a_{n}\right)=2 n$, $a_{n} b_{n}=n \times 2^{n}$.
Therefore, $T_{n}=1 \times 2^{1}+2 \times 2^{2}+\cdots+$
$$
\begin{array}{c}
(n-1) \times 2^{n-1}+n \times 2^{n} \\
\Rightarrow 2 T_{n}=1 \times 2^{2}+2 \times 2^{3}+\cdots+ \\
(n-1) \times 2^{n}+n \times 2^{n+1} .
\end{array}
$$

Subtracting the above two equations, we get
$$
\begin{array}{l}
-T_{n}=2^{1}+2^{2}+\cdots+2^{n-1}+2^{n}-n \times 2^{n+1} \\
\Rightarrow T_{n}=(n-1) \times 2^{n+1}+2 .
\end{array}
$$

Thus, the sum of the first $n$ terms of the sequence $\left\{a_{n} b_{n}\right\}$ is
$$
T_{n}=(n-1) \times 2^{n+1}+2\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(2) Notice that,
$$
\begin{aligned}
k & \leqslant \frac{1}{\sqrt{n+1}} \cdot \frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdots \cdots \frac{1+b_{n}}{b_{n}} \\
& =\frac{1}{\sqrt{n+1}} \cdot \frac{1+2}{2} \cdot \frac{1+4}{4} \cdots \cdots \frac{1+2 n}{2 n} .
\end{aligned}
$$

Construct
$$
f(n)=\frac{1}{\sqrt{n+1}} \cdot \frac{1+2}{2} \cdot \frac{1+4}{4} \cdots \cdot \frac{1+2 n}{2 n} .
$$

Then $\frac{f(n+1)}{f(n)}=\frac{\sqrt{n+1}}{\sqrt{n+2}} \cdot \frac{1+2(n+1)}{2(n+1)}$
$$
=\sqrt{\frac{4 n^{2}+12 n+9}{4 n^{2}+12 n+8}}>1 \text {. }
$$

Thus, the sequence $\{f(n)\}$ is strictly monotonically increasing.
Therefore, the minimum value of $f(n)$ is $f(1)=\frac{3 \sqrt{2}}{4}$, which means the maximum value of the real number $k$ is $\frac{3 \sqrt{2}}{4}$.