Since T1 collector is directly connected to T2 base, their voltages are the same and T2 emitter is about 0.6V lower than that. Ie2 = Ve2/Re2 ≈ Ic2. Choose Rc2 so that the voltage drop Ic2*Rc2 from Vcc will leave enough Vce for proper transistor operation.
Let's take an example. Say Vcc = 12V and the biasing of stage 1 is such that Vc1 = 2.6V. Then Ve2 will be about 2V. If Re2 = 1k, then Ie2 = 2mA ≈ Ic2. If Rc2 = 2.7k, the voltage drop = 5.4V and Vc2 = 6.6V and Vce2 = 4.6V.
We have already chosen Vc1 = 2.6V. If Rc1 = 10k, then Ic1 = (12-2.6)V/10k = 0.94mA ≈ Ie1.
If Re1 = 470Ω, then Ve1 = 0.4418V. Using 0.6V for Vbe1, Vb1 = 1.0418V.
If R2 = 10k, the current through it = 104.18µA
Add about 4µA for Ib1 (for beta ~250), then the current through R1 = 108.18µA
The voltage drop across R1 = Vcc - Vb1 = (12 - 1.0418)V = 10.9582V
R1 = 10.9582V/108.18µA = 101.3k, say 100k.
---------- Post added at 14:54 ---------- Previous post was at 14:39 ----------
The design example above should be stable enough. It's just one possible combination of bias and component values and other combinations can be used. As mentioned before, bypassing Re2 is important to have enough open-loop gain. Splitting Re1 into, say 390 + 82 ohms instead of a single 470 ohms and bypassing the 390 ohms will also result in more open-loop gain and leave a wider margin for the negative feedback factor.
You can have an improved bias stability if the upper terminal of R1 is fed from T2 emitter instead of Vcc. For that, bypassing Re2 is a must. The value of R1 will also have to be changed since it will now be fed from a lower voltage.