Answer
$$A=12\pi$$
Work Step by Step
$x^2+y^2=4$
Is an easy to guess type of equation of circle. A general equation has a form of $x^2+y^2=r^2$. So we can assume that the first circle has radius of $\sqrt4=2$ ($r_1=2$) and with center of $O(0,0)$.
But the second equation $x^2+y^2-4y-12=0$ isn't giving us anything about the circle, yet. We have to modify/simplify it the way that it will have form of a general circle equation.
$x^2+y^2-4y=12$ //Add 4
$x^2+(y^2-4y+4)=16$ //And factor out the inner side of the equation (involving $y$)
$x^2+(y-2)(y+2)=16$ //Simplify
$x^2+(y-2)^2=16$
Now we see, that the second circle has a radius of $\sqrt{16}=4$ ($r_2=4$) and it is moved upwards by $2$ units. (You can also see the image below).
So, we have to find the area of second circle ($A_2$) and subtract the first circle area ($A_1$).
$A_2 = \pi \times r^2=16\pi$
$A_1= \pi \times r^2 = 4\pi$
$A=A_2-A_1=16\pi - 4\pi = 12\pi$
