One of the problems I was solving states that "By Gauss’ law the electric field inside a sphere of uniformly distributed charges is 0". I don't understand how that is possible. I thought that this was only true for conductors. Can somone explaine why this is the case?
This is not true. The Electric Field inside a sphere with a constant volume density $\rho$ increases linearly inside the sphere, and so is only zero at the origin. This is a pretty standard derivation using Gauss's law, so I won't show it here, but the electric field for such a charged non-conducting sphere would be:
$$\mathbf{E} = \Large{\begin{cases}\frac{\rho}{3 \epsilon_0}r\,\,\hat{\mathbf{r}}, \quad & r\leq R\\ \frac{\rho R^3}{\epsilon_9} \frac{1}{r^2} \hat{\mathbf{r}}\quad & r> R\end{cases}}$$
Of course, if the sphere was conducting (but solid) or was instead a hollow shell (either a thick shell with some volume charge density $\rho$, or a "thin" shell with some surface charge density $\sigma$) then the answer you quote is correct: in both the above cases (for very similar reasons) the electric field inside the sphere is zero. (Again, this is something that can be proved almost trivially by invoking Gauss's Law and computing the net enclosed charge. In the case of the conducting sphere, you need to remember where the charge on a conductor resides.)
EDIT: The OP was changed to include the question.
In the question, you are given a section of a hollow sphere, and a point charged placed within its inner radius. The question asks you to add a charge distribution that leads to no force being exerted on the point charge. The solution does this by "completing" the hollow sphere: in this case, the entire point charge is surrounded by the hollow sphere. Since the sphere is hollow, and the charge is now "inside" it, there is no electric field and consequently no force. (The problem then goes on to deal with some strange constraint etc. but that's beside the point.)
