Small Signal BJT Model

^{simulate this circuit – Schematic created using CircuitLab}

The model of each transistor is not dependent on what it will be connected.

Yes it is possible, but you must have into account the sign of the currents and which voltage controls the collector source current, Vbe or Veb.

You can take the NPN model and apply Vbe = -Veb (that's by definition). Then, on the equations for the NPN model you'll have ib = -Veb/rb, ic = -gm Veb and ie = ib + ie;

But now all currents are negative and the arrows pointing against the flow. Since that's not intuitive and we're free to choose the signs, we consider positive the currents that flow outwards and turn around the current source. That translates to the equations to multiply the right members of the equations by -1. This way, you have the PNP model.

As you see, It's a matter of avoiding minus signs. Just like in large signal analysis we say VEB = 0.6V instead VBE = -0.6V for a PNP transistor.

The current supplied 2nd source \$g_{m2}v_{\pi 2}\$ is not negligible and is not ignored here.

The current entering to node \$v_{b2}\$ from left is the base current to second transistor (let it be \$i_{b2}\$). Then the collector current will be \$\beta_2i_{b2}\$ hence the emitter current will be \$(\beta_2 + 1)i_{b2}\$. That is why the final equation has a multiplication term of \$(\beta_2 + 1)\$ in it.

OR the 2nd source provides a current of \$\beta_2i_{b2}\$.