13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$

for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$.
Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$

From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when $x=\frac{\pi}{2}$, $f(x)$ has a minimum value of 2;
when $x=\frac{\pi}{6}$, $f(x)$ has a maximum value of 3.
(2) From the given conditions, for any $x \in \mathbf{R}$,
$$
\begin{array}{l}
2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}-c\right)+a+b=1 \\
\Rightarrow 2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}\right) \cdot \cos c- \\
2 b \cos \left(x+\frac{\pi}{3}\right) \cdot \sin c+(a+b-1)=0 \\
\Rightarrow 2(a+b \cos c) \cdot \sin \left(x+\frac{\pi}{3}\right)- \\
2 b \sin c \cdot \cos \left(x+\frac{\pi}{3}\right)+(a+b-1)=0 . \\
\text { Hence }\left\{\begin{array}{l}
a+b \cos c=0, \\
\sin c=0, \\
a+b-1=0 .
\end{array}\right.
\end{array}
$$

From $b \sin c=0$, we know $b=0$ or $\sin c=0$.
If $b=0$, then from $a+b \cos c=0$, we know $a=0$, which contradicts $a+b-1=0$.
If $\sin c=0$, then $\cos c=1$ (discard), $\cos c=-1$.
Solving gives $a=b=\frac{1}{2}, c=(2 k+1) \pi$.
Thus, $\frac{b \cos c}{a}=-1$.