13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$,
(I) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(II) If real numbers $a, b, c$ satisfy $a f(x)+b f(x-c)=1$ for any $x \in \mathbb{R}$, find the value of $\frac{b \cos c}{a}$.
13. Solution: (I) From the given, $f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1$,
From $0 \leq x \leq \frac{\pi}{2}$, we know $\frac{\pi}{3} \leq x+\frac{\pi}{3} \leq \frac{5 \pi}{6}$, thus $\frac{1}{2} \leq \sin \left(x+\frac{\pi}{3}\right) \leq 1$.
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 $2 \times 1+1=3$.
(II) From the given, we know
$2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}-c\right)+a+b=1$ holds for any $x \in R$,
$\therefore 2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}\right) \cdot \cos c-2 b \cos \left(x+\frac{\pi}{3}\right) \cdot \sin c+(a+b-1)=0$
$\therefore 2(a+b \cos c) \cdot \sin \left(x+\frac{\pi}{3}\right)-2 b \sin c \cdot \cos \left(x+\frac{\pi}{3}\right)+(a+b-1)=0$
$\therefore\left\{\begin{array}{l}a+b \cos c=0 \\ b \sin c=0 \\ a+b-1=0\end{array}\right.$,
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, $\frac{b \cos c}{a}=-1$.
(20 points)