For what value of $a$ does the equation $|x(x-4)|=a$ have exactly 3 real solutions? I simplified the equation and got two cases where $$x(x-4)=a$$
and 
$$x(x-4)=-a.$$
How do I go after that. I know how you could get $2$ solutions or $4$ solutions, but how do you get $3$ solutions?
Note that we must have $a\geq0$ or the original equation is unsolvable. We then, as you showed have

$$x(x-4)=a$$

and

$$x(x-4)=-a$$

Solving these with the quadratic formula gives

$$x=\frac{4\pm\sqrt{16+4a}}{2}$$
$$x=\frac{4\pm\sqrt{16-4a}}{2}$$

Note that, since $a\geq0$, the first equation always has two solutions. Is there then an $a$ that will make the second only have $1$? You should see that happens when $\sqrt{16-4a}=0$, which tells us $a=4$.
Hint:
Linear equations have one solution
Quadratic equations have two solutions 
Hence cubic equations will have 3 solutions