4. $4 \pi(2-\sqrt{3})$.

Connect each vertex with the orthocenter and extend to intersect the opposite sides $B C, C A, A B$ at $D, E, F$ respectively, then $A D \perp B C, B E \perp$ $A C, \angle A H B + \angle C = \angle E H D + \angle C = 180^{\circ}$, so $\frac{A B}{\sin \angle A H B} = \frac{A B}{\sin \angle A C B}$. Therefore, the circumradius of $\triangle A B H$ and $\triangle A B C$ are equal, $R = \frac{B C}{2 \sin A} = \frac{2}{2 \sin 75^{\circ}} = \frac{1}{\sin 75^{\circ}} = \sqrt{6} - \sqrt{2}$. The required area is $\pi R^{2} = \pi (\sqrt{6} - \sqrt{2})^{2} = 4 \pi (2 - \sqrt{3})$.
4. Given $H$ is the orthocenter of $\triangle A B C$, $\angle A=75^{\circ}, B C=2$, the area of the circumcircle of $\triangle A B H$ is . $\qquad$