Why is Q1 taken as 'on' and Q2 as 'off' in the differential amplifier?
Why can the voltage at the emitter not be -0.7 V so that both the
transistors would be on?

If that happened then there would be 1.7 volts across the base-emitter junction of Q1 and, given that the base-emitter junction is a forward biased diode, having 1.7 volts across it (for even a fraction of a second) would destroy the transistor: -