do not forgot that with capacitors i=(1/C)Cdv/dt. so you could say say that there will be a Ii when Vi changes.
io does not go through the capacitor because it is an open circuit on the other side, so the current has nowhere to flow.
This is not important for you solving the problem but it might help prospective. the purpose of circuits like this are to amplify an AC signal. The C1 is in there to block DC gain from entering the amplifier. Rb, Re and Rc will bias your transistor so it is operating in the correct region. C2 will prevent a DC voltage from leaving the amplifier. Since we are dealing with AC gains, all components will affect the gain. also note that in many cases circuits like this are attached to a high impedance circuit on the output, so it can be modeled as an op circuit.