Looks a lot like homework. Show us your best attempt at solving this and we will guide you on what you have correct and what you have wrong. We won't do your homework for you. You won't learn anything that way.

 

Looks a lot like homework. Show us your best attempt at solving this and we will guide you on what you have correct and what you have wrong. We won't do your homework for you. You won't learn anything that way.

I dont understand why you can't believe that I have already graduated and that I am solving questions from a work book (which supposedly contains "tricky questions") to improve my knowledge on this subject.

The question is to find both Req and Ceq. This is a solved example.
According to the solution,

Req = R1 + R2 = 6 ohms (this is completely not what I learnt about calculating Thevenin's resistance. You are supposed to short the voltage source and open current source. After sometime, capacitor acts like a voltage source and hence Req = R1 parallel R2 when all the capacitors are shorted.)
Ceq = (C1 parallel C2) series C3 which comes to 2/3 F

I am not convinced with the given solution. Since I graduated, I can't ask my teachers whom I have no longer contact with now. Which is why I posted it here.

 

I aprtially agree with your solution, 3Ω and 2/3F should be correct.

 

At DC, Req = 3.

ak

 

Looks like a trick question to me.

The reactance Xc of a capacitor is 1/2πfC..
At f = 0, Xc = ∞

Hence, remove all C from the circuit. What remains?