You have an unloaded CE amplifier.
DC Biasing gives you \$I_E\$, which allows you to determine \$r_e\$.
You would probably be given a Table for input impedance \$Z_i\$, output impedance \$Z_o\$, voltage gain \$A_V\$ and current gain \$A_I\$ for the variety of biasing circuits your instructor expects you to cover.
In the CE \$r_e\$ model, there is no connection between input and output. So Input impedance is: $$Z_i = R_B \parallel \beta r_e$$
Output impedance looks back into circuit with 0 input.
$$Z_o = r_o \parallel R_C$$
If \$r_o\$ >> \$R_C\$ (by a factor of 10+) then output impedance can be simplified to:
$$Z_o = R_C$$$$Z_o \approx R_C$$
This simplification can be made because the Q-Point will vary with \$\beta\$, temperature, tolerances.
Voltage Gain (from a formula sheet).
$$ A_v = - \frac {r_o \parallel R_C} {r_e} $$ $$ A_v = - \frac {R_C} {r_e} ,\ if\ r_o >> R_C$$$$ A_v \approx - \frac {R_C} {r_e} ,\ if\ r_o >> R_C$$
You can use \$Z_i\$, \$Z_o\$, and \$A_V\$ to form the Generalized Model to simplify solving currents (but that is not your question).
Use current division with the \$r_e\$ model to solve for currents.
$$ I_b = I_i\ \frac {R_B} {R_B + \beta r_e} $$
That gives you \$\beta I_b\$.
\$I_o = \beta I_b\$\$I_o \approx \beta I_b\$ or:
$$ I_o = \beta I_b\ \frac {r_o} {R_C + r_o} $$