A Right Angle Inside a Right Triangle

As shown in the figure, in triangle $ABC$ we have $\angle ABC = 90^\circ$, $AB = \sqrt{3}$, and $BC = 1$. The point $P$ lies inside $\triangle ABC$ and satisfies $\angle BPC = 90^\circ$.

1. If $PB = \dfrac{1}{2}$, find $PA$. 2. If $\angle APB = 150^\circ$, find $\tan \angle PBA$.