smc

The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$. (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.) The area of $AFB$ is $[AFB]=\frac{AF\cdot FB}{2}=4$, and the area of $ABCDE$ is $CD\cdot DE-[AFB]=4\cdot (2\pi +1)-4=8\pi$. From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2+4^2}=2\sqrt{5}$, thus the radius of the circle is $\sqrt{5}$, and the area of the half-circle that is inside $ABCDE$ is $\frac{5\pi }{2}$. Therefore the probability that $APB$ is obtuse is $\frac{\frac{5\pi }{2}}{8\pi }=\boxed{\frac{5}{16}}$.