Explanation

(a)
$\sin{A + \sqrt{3}\cos{A = \sqrt{3}\cos{B - \sin B}}}$
$\Rightarrow \ \frac{1}{2}\sin{A + \frac{\sqrt{3}}{2}\cos{A = \frac{\sqrt{3}}{2}\cos{B - \frac{1}{2}\sin B}}}$
$\Rightarrow \cos{\frac{\pi}{3}\sin{A + \sin{\frac{\pi}{3}\cos{A = \sin{\frac{\pi}{3}\cos{B - \cos{\frac{\pi}{3}\sin B}}}}}}}$
$\Rightarrow \sin{\left( A + \frac{\pi}{3} \right) = \sin\left( \frac{\pi}{3} - B \right)}$
$\Rightarrow A + \frac{\pi}{3} = \frac{\pi}{3} - B$
$\Rightarrow \ \ A = - B$
Now, $\sin{3(A) + \sin{3B = \sin{( - 3B) + \sin{3B}}}}$
$= - \sin{3B + \sin{3B\text{=}0}}$