Electronic – Two-stages class AB amplifier

Without more context, I can't answer the question of why different expressions are used but do note that

$$r_e = \frac{1}{g_m}||r_{\pi}$$

so the expressions are, in fact, equivalent.

To see this, recall

$$r_{\pi}= \frac{\beta}{g_m}$$

Thus,

$$\frac{1}{g_m}||r_{\pi} = \frac{r_{\pi}}{\beta}||r_{\pi} = \frac{r_{\pi}}{1 + \beta} = r_e$$

First question: No, the circuit isn't perfectly symmetrical. The current mirror performs a differential- to single-ended conversion. If you wanted a perfectly symmetrical circuit, you would make M3 and M4 current sources, and then use some sort of common-mode feedback to set the appropriate current (so that \$V_{O1}\$ and \$V_{O2}\$ stay in a usable range).

The way it is now, \$A'_{v,dm}=\frac{g_m}{2g_{md}}\$ (where \$g_{md}\$ is the transconductance of the diode-connected device M3), while \$A''_{v,dm}=g_mr_{out}\$. Remember that the impedance seen looking into M3 is simply \$\frac{1}{g_{md}}\$. These two gains are obviously very different: \$A'_{v,dm}\ll A''_{v,dm}\$. The gain \$A'_{v,dm}\$ is so small that it's pretty useless, so the signal \$v_{o1}\$ is usually ignored, and a single, single-ended output on \$v_{o2}\$ is used.

Second question: I already said it, but \$A'_{v,dm}\$ is generally so small that \$v_{o1}\$ is ignored and isn't fed to the next gain stage (or output, or whatever follows). This means that the gain is simply \$A_{v,dm}=A''_{v,dm}\$. If you really want to, though, you can take both \$v_{o1}\$ and \$v_{o2}\$ as outputs to get \$A_{v,dm}=A'_{v,dm}+A''_{v,dm}\$.