(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 $\frac{b \cos c}{a}$.
13 (1) From the given conditions, we know that $f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1$. Given $0 \leqslant x \leqslant \frac{\pi}{2}$, we have $\frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$, 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 $2 \times \frac{1}{2}+1=2$;

When $x=\frac{\pi}{6}$, $f(x)$ has a maximum value of $2 \times 1+1=3$.
(2) From the given conditions, we know that $2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}-c\right)+a+b=1$ holds for any $x \in \mathbf{R}$, so
$$
\begin{array}{c}
2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}\right) \cos c- \\
2 b \cos \left(x+\frac{\pi}{3}\right) \sin c+(a+b-1)=0, \\
2(a+b \cos c) \sin \left(x+\frac{\pi}{3}\right)-2 b \sin c \cos \left(x+\frac{\pi}{3}\right)+(a+b-1)=0,
\end{array}
$$

Therefore, $\left\{\begin{array}{l}a+b \cos c=0, \\ b \sin c=0, \\ a+b-1=0,\end{array}\right.$ 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 $a=b=\frac{1}{2}$, $c=(2 k+1) \pi$, so, $\frac{b \cos c}{a}=-1$.