Digital Circuits Analysis
First of all , when Q4 is ON  and Q3 is OFF (so it is a open circuit ideally) , we can use thevenin equivalent at the emitter node of Q1 ,which gives you VTh= 2.5V and Rth =0.5k. under the assumption that R1 is connected to 5V 

Now applying KVL at the input loop from 5V input to the ground node,we have

5 - (10K) Ib  - Vbe - Ie (0.5K) -2.5 = 0

Rearranging,

1.8V  = (10K) Ib +  (0.5k) Ie

but Ie =(1 + beta) Ib;

So,

1.8V  = [ (10k /51) + 0.5k ] Ie

which gives Ie = 1.8/0.696k  =2.585 amps.

edit: and regarding 10v 2.5v and 5v confusion, think of it like this,

we have got resistive divider with supply voltage 5V at R1 which is in series with R2 giving us 2.5 volts at emitter node.

The 10v used in the circuit seems vague as the supply itself is 5V,so thats wrong i guess.