How were values of Vc(0+) and i(1.5ms) found?
When the switch is open, the capacitor will charge to whatever voltage appears across the 200 ohm resistor.

Once the capacitor charges, all the current from the 100mA source will go through the 200 ohm resistor, and the voltage that'll appear across the resistor will be:

$$ E= IR =0.1A \times 200\Omega = 20 \text{volts}$$ 

Then, when the switch closes, the 100mA from the current source will be shunted to ground through the switch as will the current from the cap as it discharges through 50 ohms to ground, the 200 ohm resistor being shunted by the switch.