Determine the voltage drops across the capacitor
I think you have the first part (switch closed) right.  Then, the top R, the two R's in parallel, and the final R all form a voltaage divider with the C at that voltage.

When you open the switch, however, the C will charge to the full E; the current flowing through the two resistors now in 'parallel' with the C don't matter.  However, the rate at which it charges will be determined by all the resistors in series with C.  

And, as usual, the solution will look like 

$$V_0+ V_fe^{-f(r,c)t}$$