Looking into the base terminal we see the equivalent of a resistor of value Re *hfe, so if hfe is 200, it looks like a 1.5M resistor to ground.
They are saying we can ignore that if R1 || R2 << (Re * hfe), where they consider an order of magnitude to be close enough- so a reduction in swing of Vcc/20 is considered insignificant. There's nothing stopping you from correcting the ratio a bit to account for typical hfe, but when AoE was written 5% resistors were much cheaper than 1% and it didn't matter that much.
If you look at CB you'll see that it decouples Q1 base, i.e. provides a reference voltage. So this IS just a classic cascode.
Rule of thumb Ve(Q2) = Vcc/10 is fine. That gives you Vb(Q2). And Ie(Q2), Ib(Q2) from Re and Hfe.
Now Q2 works into a low impedance (Q1e) so Vc varies very little and Q2 Vce can be small. I'd suggest 2 or 3x Vbe or a couple of volts. If Vcc is low and you need high output range, reduce it a bit : if linearity is important, remember hfe reduces as Vce gets close to saturation, so increase it a little, or set Vc(Q2) to Vcc/3.
That gives you Ve(Q1), Vb(Q1), Ib(Q1) since Ie(Q1) = Ic(Q1) = (Ie-Ib)(Q1), so you can now fix the bias chain.
Now split the difference between Ve(Q1) and Vcc, to give you Vc(Q1) for maximum symmetrical output swing. That and Ic(Q2) gives you RC.
Best Answer
It is an approximation.
You typically work around the approximation by setting the current flow the two bias divider resistors be 10X what the maximum expected base current would be. Some folks like to use a higher ratio than 10X.