在 $\triangle A B C$ 中, 角 $A, B, C$ 对应边分别为 $a, b, c$, 若 $\sqrt{3} a \sin C+a \cos C=c+b$.

(1) 求角 $A$;

(2) 若 $a=\sqrt{3}$, 求 $b+c$ 的取值范围.
(1) $\because a \cos C+\sqrt{3} a \sin C=b+c$, $\therefore$ 由正弦定理可得 $\sin A \cos C+\sqrt{3} \sin A \sin C=\sin B+\sin C$, $\therefore \sin A \cos C+\sqrt{3} \sin A \sin C=\sin (A+C)+\sin C, \therefore \sqrt{3} \sin A-\cos A=1, \therefore \sin \left(A-30^{\circ}\right)=\frac{1}{2}, \therefore A-30^{\circ}=30^{\circ}$, $\therefore A=60^{\circ}$;

(2) 由题意, $b>0, c>0, b+c>a=\sqrt{3}, \therefore$ 由余弦定理 $3=b^{2}+c^{2}-2 b c \cos 60^{\circ}=(b+c)^{2}-3 b c \geq \frac{1}{4}$ $(b+c)^{2}$ (当且仅当 $b=c$ 时取等号), 即 $(b+c)^{2} \leq 12, \therefore b+c \leq 2 \sqrt{3} . \because b+c>\sqrt{3}, \therefore \sqrt{3}<b+c \leq 2 \sqrt{3}$.