Problem 203

Solution:
(1) Since $\overrightarrow{m}=(\sqrt{3}\sin x,2)$ and $\overrightarrow{n}=(2\cos x,\cos^2 x)$,
then $f(x)=\overrightarrow{m}\cdot\overrightarrow{n}=2\sqrt{3}\sin x\cos x+2\cos^2 x=\sqrt{3}\sin 2x+\cos 2x+1=2\sin(2x+\frac{\pi}{6})+1$,
Since $-1\leqslant\sin(2x+\frac{\pi}{6})\leqslant 1$,
then $-1\leqslant 2\sin(2x+\frac{\pi}{6})+1\leqslant 3$,
thus, the range of the function $f(x)$ is $\boxed{[-1,3]}$;
(2) Since $f(A)=2$,
then $2\sin(2A+\frac{\pi}{6})+1=2$,
thus $\sin(2A+\frac{\pi}{6})=\frac{1}{2}$
thus $2A+\frac{\pi}{6}=2k\pi+\frac{\pi}{6}$, or $2A+\frac{\pi}{6}=2k\pi+\frac{5\pi}{6}$, $k\in\mathbb{Z}$,
thus $A=k\pi$ (discard this), $A=k\pi+\frac{\pi}{3}$, $k\in\mathbb{Z}$,
Since $0 < A < \pi$,
thus $A=\boxed{\frac{\pi}{3}}$,
Since $\sin B=2\sin C$, by the Law of Sines, we get $b=2c$,
Since $a=2$, by the Law of Cosines, we get $a^2=b^2+c^2-2bc\cos A$,
thus $3c^2=4$,
solving this, we get $c=\boxed{\frac{2\sqrt{3}}{3}}$.