Determining the state of this BJT transistor
Your first explanation, starting with Q1 is correctly describing the state of the circuit.

But if I initially start thinking from Q2 then I can say that initially Vin2=-1V and Vee=-9V, hence Vbe2 is forward biased and the transistor Q2 is active.

This is indeed incorrect as it ignores the fact that a current is coming out of the emitter of Q1. So much current that the voltage across RE becomes more than the 7.3 V (or less) needed to keep the Base-Emitter junction of Q2 in forward mode.

Your explanation would be correct only if there was no current coming out of Q1. But there is and so much in fact that the common emitter rail (point A) is pulled up so much that Q2 ends up being cut off.