I took the capacitor as an open and did a KVL from the 2V sourece to the 1.2kohm.
-Vb+Vbe+Ie(100+1.2k)=0
Ie=(2-.70/(100+1.2k)=1mA
Ie is approximate to Ic, Ic=1mA
Then a KVL at the output
-10V+1k x Ic+Vout=0
Vout=9V
The only thing I know for AC analysis is re=25mV/Ie=25ohm
Are you able to draw the small signal AC equivalent circuit? This is essentially a grounded base configuration. How would you know that it is grounded base?