For the circuit you've shown us, a CC arranged topology with a bipolar transistor, the input capacitor is usually taken as a wire (for AC purposes) and isn't counted in the input impedance.
(Technically, it's impedance is orthogonal to the resistive elements and usually doesn't significantly modify the resulting impedance.)
The AC signal source for your circuit is taken to be single-ended (since the bipolar transistor only has 3 wires, the signal source must be single-ended.) So the AC signal source applies itself to the base, with reference to either ground or the supply rail.
input impedance
An AC signal source will see its own source impedance (not shown, so here taken to be zero), this then added to what your circuit shows:
- \$R_1\$ has one end tied to a voltage source. So the signal source sees \$R_1\$ as a load.
- \$R_2\$ has one end tied to a voltage source (ground qualifies, too.) So the signal source sees \$R_2\$ as a load.
- The bipolar transistor device's base is part of a 3-wire device that includes an emitter and collector, as well. Slight voltage variations at the transistor base will necessarily involve slight current variations with respect to the DC quiescent base current. This is effectively yet another resistance. The problem is how to work out the exact current change for an exact voltage change at the base.
Taken all together, the above three resistances are in parallel with each other as all of them experience a current change and the signal source makes a voltage change.
The idea above also includes only making this measurement around the DC quiescent point and to only use an infinitesimally small voltage change (tiny) and to see what current compliance is required by the signal source when making that tiny voltage change. So this is a small signal analysis. Not a large signal one.
If the base is lifted up by a tiny \$\Delta v_{_\text{IN}}=\Delta v_{_\text{B}}\$, this incurs similar change at the emitter. That change at the emitter causes an emitter current change of \$\Delta i_{_\text{E}}=\frac{\Delta v_{_\text{IN}}}{R_{_\text{E}}}\$. This causes a change in the base current of \$\Delta i_{_\text{IN}}=\Delta i_{_\text{B}}=\frac{\Delta i_{_\text{E}}}{\beta_{\small{F}}+1}\$. So \$R_{_\text{E}}\$'s effective resistance seen at the base is \$\frac{\Delta v_{_\text{B}}}{\Delta i_{_\text{B}}}=R_{_\text{E}}\left(\beta_{\small{F}}+1\right)\$. In addition, there's the dynamic impedance due to Ebers-Moll of \$r_\pi=\frac{\beta_{\small{F}}}{g_m}\$(see here).
So the answer for the input impedance, as seen by the signal source, is:$$R_i=R_1\:\vert\vert\: R_2\:\vert\vert\: \left[r_\pi+R_{_\text{E}}\left(\beta_{\small{F}}+1\right)\right]=R_1\:\vert\vert\: R_2\:\vert\vert\: \left[\left(r_e^{\:'}+R_{_\text{E}}\right)\left(\beta_{\small{F}}+1\right)\right]$$
output impedance
This asks what \$\Delta i_{_\text{OUT}}=\Delta i_{_\text{E}}\$ results from \$\Delta v_{_\text{OUT}}=\Delta v_{_\text{E}}\$:
- \$R_{_\text{E}}\$ has one end tied to a voltage source (ground). So the change sees \$R_{_\text{E}}\$ as a load.
- The emitter is also seen as a load. This also has to be worked out.
This question, though, is a little trickier. The reason is that there is usually a source impedance that isn't zero, usually just called \$Z_{_\text{IN}}\$. This isn't shown in your schematic and in the prior section I took \$Z_{_\text{IN}}=0\:\Omega\$ for analysis. In this case, if we again assume \$Z_{_\text{IN}}=0\:\Omega\$, then the base itself looks like an ideal voltage source. As a consequence of that assumption, the values of \$R_1\$ and \$R_2\$ are nullified. So the base itself just looks like \$0\:\Omega\$ from the emitter. This leaves only the Ebers-Moll dynamic impedance of \$r_e^{\:'}=\frac{\alpha_{\small{F}}}{g_m}\$(see here, again), which is usually a lot smaller than \$R_{_\text{E}}\$.
But to complete the answer with \$Z_{_\text{IN}}=0\:\Omega\$, then the output impedance is:
$$R_o=R_{_\text{E}}\:\vert\vert\:r_e^{\:'}=R_{_\text{E}}\:\vert\vert\:\frac{r_\pi}{\beta_{\small{F}}+1}$$
If \$Z_{_\text{IN}}\gt 0\:\Omega\$ then the output impedance is:
$$R_o=R_{_\text{E}}\:\vert\vert\:\left[r_e^{\:'}+\frac{R_1\:\vert\vert\: R_2\:\vert\vert\: Z_{_\text{IN}}}{\beta_{\small{F}}+1}\right]=R_{_\text{E}}\:\vert\vert\:\frac{r_\pi+R_1\:\vert\vert\: R_2\:\vert\vert\: Z_{_\text{IN}}}{\beta_{\small{F}}+1}$$
since \$r_e^{\:'}\$ is in series with the emitter-reflected base impedances.
(The above excludes the Early Effect.)