Voltage divider, transistor question
No, that won't work. By adding Q1 you have dragged up the R2/R3 node to 0.7 volts below Vcc. The base-emitter junction of the BJT is a forward biased diode and it will restrict the voltage to about 0.7 volts.

On the other hand this forward biased diode is fighting two forward biased diodes inside Q3. You will have the Vcc rail dragged down to about 2 or 3 volts and, if that rail can supply enough current you will destroy either Q1 or Q3.