Which equation of the Ebers-Moll-Model is correct

αF is not the Hfe of the transistor. Per the model, αF of the emitter current reaches the collector.

This means that αF of the diode current passing through the base-emitter junction contributes to the current flowing through the base-collector junction.

Typically, αF has a value of between 0.98 and 0.99. the forward beta (~hfe) is αF/(1-αF).

The first thing you should notice is that from a small-signal preceptive point of view \$Q_1\$ \$r_{o1}\$ act just like the emitter resistance for \$Q_2\$

And the small signal mode will look like this

^{simulate this circuit – Schematic created using CircuitLab}

The small-signal parameters are:

\$\large g_{m2} = \frac{0.1\textrm{mA}}{25\textrm{mV}} = 4\textrm{mS}\$

\$\large r_{\pi 2} = \frac{\beta}{g_{m2}} = 25\textrm{k}\Omega\$

\$\large r_{o2} = r_{o1} = \frac{V_A}{\textrm{I}} =\frac{5 \textrm{V}}{0.1\textrm{mA}} = 50 \$

Hence the output resistance will be the same as I show here:

BJT common-base output resistance derivation

finding output resistance of CB amplifier with ro

$$R_{OUT} = \frac{V_X}{I_X} = r_{o2}+\left(1+ g_{m2} r_{o2}\right)\cdot r_{o1}||r_{\pi2} = 3.4\textrm{M}\Omega $$

Also, remember that in the small-signal analysis you can replace PNP with the NPN small-signal model.

^{simulate this circuit}

^{simulate this circuit}