For question 1:
First let's start with the direction of the field. It is important to know that field lines point the way a (+) charge would migrate (see the top of page 121 for the details). In order to suspend a negatively charged oil drop, you need to have the electric force on the electron-rich oil drop pointing up to offset the force of gravity pointing down. So the electric force on the negative charge is up, which would mean that that same force would point downward if it were a (+) charge. Because field lines point the way the force would point on a (+) charge, the field lines must be pointing downward. This eliminates choices C and D.
As for the magnitude, you know that qE = mg, which tells us that E = mg/q. The mg term must be in the numerator, so choice A is the only choice still standing. Just to be complete, rho is charge density in this question, so charge density x volume gives charge. The total charge is 4/3πr^3 x (rho), so the answer is mg/[4/3πr^3(rho)] = 3mg/4πr^3(rho).
