Calculation of \$V_o/V_i\$
"Can anyone give some tricks where to start etc"

OK - here is a rough description of my approach.


At first calculate the gain Ao from base to collector using the known formula transconductance x resistance Rtot (total at the collector): Ao=-gm*Rtot. Transconductance gm=beta/Rpi.
Rtot=Rc||RL||Rbc with Miller-resistor at the collector node Rbc=Rb/(1-1/Ao), Ao negative!
Hence: Ao=-gm/(1/Yc)+gm/(1/YL)+gm*(1-1/Ao)/(1/Yb)  .
This equation must be solved for Ao (Ao appears twice!).
As the last step, you calculate the input resistance Rin=Rpi||Rbb at the base node using the Miller-Theorem (at the base) with Rbb=Rb/(1-Ao).
Finally, you can compute the resulting gain A=Ao*Rin/(Rin+Rs)


EDIT_1 (comment): Vo/Vi=Ao because Vi=V at the base node.

EDIT_2: Because you specifically ask for the exact role of Rb: 

The classical procedure to calculate the output voltage Vo leads to a negative value (with respect to Vi) because the gain Ao is inverting. However, there is another unwanted part of Vo which is NOT inverted. This part is not yet included in my above calculation. This part of the output voltage arrives DIRECTLY at the collector node (not via the transistors collector current) and is found using the simple voltaged divider rule Vi*Rtot/(Rtot+Rb). As you can see, this (positive) part disappears for Rb>>Rtot.