Pre-amplifier biased to one-half of power source
There is some interaction going on between the collector current of T2, the emitter resistor R4 and that voltage divider R2, R3.

If we assume that the current through R5 and R7 is so small that there's almost no voltage drop across them then the voltage across R3 is equal to Vbe_T2 + V_R4.

That Vbe_T2 will be quite constant at around 0.6 V.

But what will the voltage across R4 be? It depends on the current through it of course. But since we want the voltage across R2 + R3 to be Vcc/2 is means the voltage across R3 is Vcc/4. From that Vcc/4 subtract the Vbe of T2, that leaves 3 V / 4  - 0.6 V = 0.15 V across R4. So the current through R4 will be 0.15V / 1.5k ohm = 100 uA.

Then we need to check if the rest of the circuit can support the values we just calculated. At the collector of T2 we will have 3 V - 100 uA * 12 kohm = 1.8 V
 Since T2's emitter is at 0.15 V that leaves a Vce of 1.65 V which is enough.

Base of T1 is at that same 1.8 V, subtract Vbe to get to emitter voltage: 1.8 V - 0.6 V = 1.2 V That is a bit lower than the 1.5 V we wanted, so you would think that T1 would be off (Vbe is too low) and the circuit cannot work.

However due to the feedback the circuit will automatically find a point where it can work.

What happens is that when T1 is off, the voltage across R2, R3 is lower than we want and that also means that the voltage across R4 will be less. So T2 will have a much smaller Ic than we predicted. This makes the voltage at the collector of T2 rise. Which is what we need to turn T1 on. That in turn will increase the voltage across R2, R3 and R4 until a certain balance situation is found where the voltages and currents will be close to what we predicted.

You could re-do the calculation using a larger Vbe, for example Vbe = 0.65 V instead of 0.6 V, that would make Ic of T2 a bit smaller and decrease the voltage across R1 (which was the issue we had above) then perhaps you get closer to the real solution.

When designing such a circuit I do the hand calculation as explained above and then use a circuit simulator to find the exact values and check if my calculation was sane (close enough I mean).