voltage and current after closing the switch
You have the right general idea, but you can't just consider the two capacitors as one 3F capacitor.  Just before the switch is closed, the 2F capacitor will be fully charged and (I presume) the 1F capacitor is fully discharged.  So when the switch is closed, the 2F capacitor will discharge and the 1F capacitor will charge.

Remember that \$Q=CV\$ for a capacitor, and that \$\displaystyle \frac{dQ}{dt} = C\frac{dV}{dt} = i_C\$ and this should help you derive the answer.