Transistor as a Switch in Cutoff and Saturation
1a) I'd say both are correct. At any voltage Vbe &lt; Vth the transistor will "cut off", where by "cut off" we mean "the current drops to a super-low value, and gets even lower as Vbe drops further". It's always an approximation -- do we say the transistor is "cut off" once the current is less than a milli-amp? Or less than a micro-amp? For a milli-amp, driving Vbe down to 0.1V might be good enough. For a micro-amp we might need to drive Vbe to 0.0V or even a negative voltage.

1b) Vbe is less than 0.7V because we have tied the input to ground. The base is "grounded" because it is tied to 0V through Rin.

2a) Not actually a question ;-)

2b) We are using Vcc as a convenient voltage to drive Vin high, so that we can drive Vbe to a nice positive voltage and turn the transistor on.