Analyzing circuit with dependent sources

As always, it's helpful to first draw the DC and AC circuits.

DC circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The operating point is evident by inspection:

$$I_C = \frac{\beta}{1 + \beta}I_2 = \alpha I_2 $$

$$V_C = I_C(\frac{75\Omega}{\alpha} + \frac{100k\Omega}{\beta}) + V_{BE} $$

Update to address comment:

I can't perfectly grasp your equation for Vcc.I think understand you divide resistance with beta and alpha to make them equivalent resistance looking through C.

Assuming you meant \$V_C\$ rather than \$V_{CC}\$, by KVL we have

$$V_C = V_E + V_{BE} + V_{R1}$$

We have

$$V_E = I_E R_S = \frac{I_C}{\alpha}R_S $$

and

$$V_{R1} = I_B R_1 = \frac{I_C}{\beta}R_1$$

Thus

$$V_C = I_C(\frac{R_S}{\alpha} + \frac{R_1}{\beta}) + V_{BE} $$

AC circuit:

^{simulate this circuit}

The small-signal circuit is thus

^{simulate this circuit}

This is a straightforward circuit to solve. What have you tried so far?

Voltage at the node where R2 is connected is Vo, consider voltage at other node is \$V_1\$ By KCL:

$$\frac{V_o}{R_2} + \frac{V_o-V_1}{R_1} + \frac{V_o-V_2-2V_o}{R_3} = 0$$

To find \$V_2\$ here it is simply \$R_4 \cdot I_1 = 48V\$.

Now put the values and you will get your answer.

However you can do same using the loop analysis and KVL.

The thing is keep the dependent source variable as it is and just solve as you normally do.