How to calculate the number of waters of hydration?
All of the questions involve the same reasoning, so let's look at question #2.

What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?