在 $\triangle A B C$ 中, $a, b, c$ 分别是角 $A . B, C$ 的对边, 且 $c=\sqrt{2,} A=105^{\circ}, C=30^{\circ}$

(1) 求 $b$ 的值

(2) $\triangle A B C$ 的面积.
(1) $\because A=105^{\circ}, C=30^{\circ}, \therefore B=45^{\circ}$, 又 $c=\sqrt{2}, \sin C=\frac{1}{2}$, $\therefore$ 由正弦定理 $\frac{b}{\sin B}=\frac{c}{\sin C}$ 得: $b=\frac{c \sin B}{\sin C}=\frac{\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{1}{2}}$ $=2$;

(2) $\because b=2, \quad c=\sqrt{2}, \quad \sin A=\sin 105^{\circ}=\sin \left(60^{\circ}+45^{\circ}\right)=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}=\frac{\sqrt{6}+1}{4}$, $\therefore S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{1}{2} \times 2 \times \sqrt{2} \times \frac{\sqrt{6}+1}{4}=\frac{1+\sqrt{3}}{4}$.