Electronic – base impedance of a bjt amplifier

The very short answer: R2 helps to turn the BJT off fast.

A bit longer: When the current ceases to be provided through R1, all there is left to stop the base from being forward biased is the current that flows through the base itself. This works for slow circuits. For fast turn-off, R2 helps to get the charge out of the base. Note that there's a parasitic capacitance from B to E and worse, from C to B. The latter is worse because while B goes down, C goes up, and the C-B-capacitor pushes charge into B while you want to lose charge in B (Miller Effect).

Also, with B floating, a tiny current (interference / parasitic creepage path) may be enough to turn the BJT on. R2 helps to prevent this.

The "gain" formula as given by you is the transconductance of a transistor in common-emitter configuration WITH feedback.

1.) Without feedback (RE=0) this transconductance is simply $$gm = \frac{hfe}{hie}$$

2.) With feedback we have $$A= \frac{gm}{1+gm*R_E}= \frac{hfe}{hie+R_E}$$

From this, we can derive the loop gain Aloop=-gm*Re. Note that gm=hfe/hie.

3.) Based on the loop gain expression (and knowing that the output resistance will be increased using feedback) we can expect (approximation):

$$Z_{out}= Z_o*(1+gm*R_E)=r_{ce}*(1+gm*R_E)$$

4.) Comment: The above expression for Zout is an approximation because it is derived under the assumption of IDEAL voltage feedback. However, the feedback voltage is provided across RE (not an ideal voltage source). More than that, also the influence of the input circuitry was neglected. However, we should not forget, that during practical operation of this circuit the large value of Zout is paralleled with the collector resitor Rc (which is much smaller that Zout). Hence, the approximation seems to be acceptable.