No, it doesn't have to be zero. Imagine that most of the charge is located on one side, or if on one side it's more to the top and the other has more to the side; the x-component wouldn't be 0.
You have an equation already; are you asking how to solve for t?
It's a quadratic equation...
EDIT: That equation doesn't have any real solutions; are you sure that 142.5 is the initial velocity in the y-direction?
The magnitude of the vector is always positive; you have the angle to account for the direction.
However, the actual vector can be going in the negative direction.
You posted Coulomb's law for the electric force. Draw a diagram and use vectors to represent the each force on the charge at the origin, then you should be able to get the magnitude of the resultant vector.
You have to go a little bit outside of plugging values into the kinematic equations.
Hint: The maximum distance the ball goes is 90m.
EDIT: Sid beat me to it.
You should have the two perpendicular force vectors starting at the origin; they form a right triangle... If you have the two legs of a right triangle, you should be fine figuring out the rest of the parts of the triangle.
For part a, you used Coulomb's law to find the force exerted by each charge on the charge at the origin. Have you paid attention to their directions? If you draw them as vectors, how would the resultant vector look?
And for electric field, you want to use the other charge as the Q you're using.
You're essentially finding the force exerted by a charge on an infinitesimally small positive charge, which happens to be placed at the origin.
electric field = kQ/r^2
flux = EA = Q/ε
You're given the electric field at both surfaces, so you can find E. A is just the area of the two squares.
Then ε is constant, just solve for Q.