What will happen when two capacitors C1 and C2 are added in series instantly by a switching mechanism, How the charge sharing will take place

Short Answer:

Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor)

Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor)

Long Answer:

A capacitors charge is given by \$Vt=V(1-e^{(-t/RC)})\$ where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.

At the exact instant power is applied, the capacitor has 0v of stored voltage and so consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.

A nice page with graphs and some math explaining this is http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_rc.html

For an inductor, the opposite is true, at the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), as time continues, it will have a low resistance to the steady voltage and carry lots of current (short circuit).

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.