It is not necessary to use a grounded emitter, but consider the alternative
simulate this circuit – Schematic created using CircuitLab
A transistor used as a switch (in saturation) will typically have a collector-emitter voltage of about 0.2 volts. Since the base-emitter voltage will be about 0.7 volts, Vs must be at least 0.5 volts above Vcc, plus whatever voltage is required across R2 to get the base current up to the level required. And that base current will be significant. Regardless of "ordinary" gain, an NPN transistor in saturation will display a much lower gain, with the typical rule of thumb being a gain of 10 to ensure low Vce. So the circuit as shown cannot be used without a second, higher power supply, which is not what you'd call convenient.
This, in turn, answers your third question. Since the transistor will be (by normal, linear standards) grossly overdriven, gain variations among transistors will typically have no obvious effect. In the circuit shown, a 50% voltage increase will cause the transistor voltage to increase from 0.2 volts to 0.3 volts, which will drop the load voltage from 4.8 to 4.7 volts, and for displays and LEDs and such this will be unnoticeable.
As to question 2, the answer is definitely yes. In many respects FETs and MOSFETs are easier to drive, since they require very little gate current (except during transitions). And, in fact, CMOS is the dominant technology for microprocessors and graphic chips, with potentially millions of transistors per chip. Well, actually, high-end CPUs and graphics ICs nowadays run between 1 and 2 billion transistors. Trying to do this with BJTs would simply be impossible due to the current requirements.