OK, investment it is. Although I'd like to see some return on investment
Just kidding.
1. Input follower
as your equation point out, Q32,Q33,X2,X3 is not only for bias , but also for load of input follower.
Yes, I agree with you that those transistors influence the loading of the followers. But if we take for example P22, it also influences the loading of Q30 and yet I hope you'll agree that it is primarily for biasing and not in the signal path. In the same way Q32-33 are used to bias Q31 Q34 and also they allow for the voltage at the emitters of X16-17 to be exactly where it should be with respect to the input common-mode voltage. Transistors Q32-33 also waste half of the signal current of the input followers, exactly because they influence the loading.
So the Gm is
2*gm1*gm2/(gm2+2gm1), then the gain of the follower is
2*gm1*gm2/(gm2+2gm1)*(gm2+2gm1))
Here, I haven't tried to verify if you are right or not simply because in my derivation I didn't care about the voltage transfer from the differential input to the emitters of the followers. I was interested in the signal current out of these followers. If we assume that the circuit is symetric, it is easy to see that the differential impedance the two input followers see at their emitters is 1/gm2. Now, if you load the input differential emitter follower structure with this differential inpedance (which means it is connected between the emitteres) and disregard 1/Rπ of the follower transistors with respect to gm1 and gm2 you'll see that the current going into the differential load of 1/gm2 is:
i=Vdi* (gm1gm2)/(gm1+2gm2)
If we assume that betta is very big and disregard the base current, half of the current above is going to the collectors of Q31 Q34 and into the load there.
2. Sencond Stage
Gain is gm2*(0.5R43||0.5(1/gm3))
I think the load of this stage is 0.5*R43 parallel with 1/gm3, can you explain your RL more clearly?
That collector current sees the load of the resistors R39, R42 and R43 and the load of the PNP transistors - note I'm talking about differential load here. For RL see the attached picture
In this case RL=(2*R43)//R39. In parallel to RL you have the load presented by the emitteres of the PNP transistors, which is 1/gm3. We are interested only in the current that goes in this 1/gm3 which is RL/(1/gm3+RL)*i. Half of this current is again wasted in X2-3, but then we add the collector currents of X1 and Q29 at the output, so for the output current we don't have to devide by 2 as before.
The output current now goes into a load which is the parallel combination of
Ro(x1), Ro(Q29) and the input impedance of Q30 (we disregard the compensation network, because we talk DC here only). The output current into this load will give you the input voltage of the output stage and the output voltage of the amplifier now is easy to find - just
gm(Q30)*(Ro(Q30)//Ro(P22)).
That's all for the investment.