Calculating capacitor current
Start with the load. ArcCos of 0.75 is 41.41 degrees. Work out the equivalent parallel R and L currents. I get I_R = 26.25 amps and no matter what you do with the capacitor this will always be 26.25 amps. I_L I get to be 23.15 amps (35 x Sin(41.41)).

What will be the net reactive current with the capacitor fitted comes next. At a PF of 0.96 the angle is 16.26 degrees. So take the Tan of 16.26 and using I_R calculate the net reactive current. I get 7.66 amps.

The capacitor current is exactly opposing (and cancelling) the inductor current so the current taken by the capacitor is 23.15 amps - 7.66 amps = 15.49 amps.

Using V, F and 15.49 amps I calculate capacitance to be 205 uF.

It all helps if you draw a phasor diagram then you can see why I have used tan and sin in various places. Let me know if you are having problems with that.