18. (12 points) In $\triangle A B C$, $a$, $b$, and $c$ are the sides opposite to the interior angles $\angle A$, $\angle B$, and $\angle C$, respectively, and satisfy $\sin A+\sqrt{3} \cos A=2$.
(1) Find the size of $\angle A$;
(2) Now, three conditions are given:
(1) $a=2$,(2) $\angle B=45^{\circ}$,(3) $c=\sqrt{3} b$.
Select two conditions that can determine $\triangle A B C$, write down your choice, and use it to find the area of $\triangle A B C$ (only one selected scheme is required).
18. (1) According to the problem, we have
$$
\sin \left(A+\frac{\pi}{3}\right)=1
$$

Since $0<\angle A<\pi$, therefore, $\angle A=\frac{\pi}{6}$.
(2) Scheme 1 selects conditions (1) and (2).

From $\frac{a}{\sin A}=\frac{b}{\sin B} \Rightarrow b=2 \sqrt{2}$.
Also, $\sin C=\sin (A+B)$
$=\sin A \cdot \cos B+\cos A \cdot \sin B$
$$
=\frac{\sqrt{2}+\sqrt{6}}{4} \text {. }
$$

Then $S=\frac{1}{2} a b \sin C$
$$
=\frac{1}{2} \times 2 \times 2 \sqrt{2} \times \frac{\sqrt{2}+\sqrt{6}}{4}=\sqrt{3}+1 \text {. }
$$

Scheme 2 selects conditions (1) and (3).
By the cosine rule, we get
$$
\begin{array}{l}
b^{2}+c^{2}-2 b c \cos A=a^{2} \\
\Rightarrow b^{2}+3 b^{2}-3 b^{2}=4 \\
\Rightarrow b=2, c=2 \sqrt{3} .
\end{array}
$$

Then $S=\frac{1}{2} b c \sin A$
$$
=\frac{1}{2} \times 2 \times 2 \sqrt{3} \times \frac{1}{2}=\sqrt{3} .
$$