Computing the gain on a small signal model
This is the typical exercise where the extra-element theorem or EET forged by Dr. Middlebrook in 1989 can be applied to obtain the response in a clear and well-ordered form - read a low-entropy form - without losing yourself in complicated KVL and KCL analysis. This is an example that I described in my book on transfer functions using the fast analytical circuits techniques or FACTs.

First, you identify the extra element or the one that bothers you in this analysis. Obviously, it is the feedback resistor \$R_1\$. We have the choice to either temporarily short it or open-circuit it. I chose to open-circuit it and determine the transfer function in this condition based on the below equivalent circuit:



If you do the maths ok, you should find: \$\frac{V_{out}}{V_{in}}=-\frac{\beta R_C}{r_{\pi}+(\beta+1)R_E}\$

Now, the exercise consists of determining two resistances \$R_d\$ and \$R_n\$ in two different conditions. For the first one, \$R_d\$, you zero the excitation source: replace \$V_{in}\$ by a short circuit and you "look" through the connecting terminals of the extra element - \$R_1\$ - to determine the resistance \$R_d\$ offered in this condition. The drawing is shown below:



Here, by inspecting the circuit, you see that there is no base current so the only resistance "seen" from the connecting terminals is \$R_C\$. We have \$R_d=R_C\$.

For the final step, we have to null the response. \$V_{in}\$ is coming back as a stimulus and a test generator \$I_T\$ is installed across the connecting terminals of \$R_1\$ to determine the resistance \$R_n\$ which nulls the output current. This is what is a called a null double injection or NDI. See the sketch here:



If you determine the resistance in these conditions, you find \$R_n=-(\frac{r_\pi}{\beta}+R_E(1+\frac{1}{\beta}))\$

Once you have these expressions on hand, you assemble them according to the EET definition: \$\frac{V_{out}}{V_{in}}=-\frac{\beta R_C}{r_{\pi}+(\beta+1)R_E}\frac{1+\frac{R_n}{R_1}}{1+\frac{R_d}{R_1}}\$.

A quick Mathcad sheet gives a numerical application:



In this example, you can see that \$R_n\$ features a negative sign. Should you replace \$R_1\$ in your circuit by a capacitor, then you would create a nice right-half-plane-zero or RHPZ.

The EET is not really complicated but it requires to go through a few examples to acquire the skill but once you have adopted the FACTs, there is no return : ) For instance, in you exercise, I can see that you are asked to determine the output and input resistances. For the output resistance, the denominator \$1+\frac{R_d}{R_1}\$ won't change and it will actually become the numerator of the input impedance transfer function. So you already have 1/3rd of the work done.