Using built-in current limiting of an OpAmp to create a current source
You need the output to be a minimum 6.8 volts (-Vs + 6.8), so you will need a resistor in series with R3 on the output as shown.  The 50 ohm load only provides 5 volts at 100 ma and your sense resistor adds 0.7 volts.  Also keep in mind that the output current limit will vary somewhat as the base-emitter voltage changes with temperature.
Using a switch as you have drawn will be ok.  Expect some spikes when you switch.


3 and 4. In addition to comment (1.) you would need the input to be a minimum of 15 volts (-VB + 15), so you will have to bias up the signal on the input op amp.


A capacitor in parallel with R2 will slow down the rise and fall times.