For the first part, I would use 9-2=7V
Then use R=V/R
So R=7/10*10^-3

R=700 ohms

For the next bit, I would draw a diode in inverse parallel to the L.E.D.

However for the last part, I am not sure how you would get the answer. Is it just half the original Resistance?

 

One question is that why a LED needs reverse protection when there is a resistor. The Led will light in either polarity of the AC.
Now if it is DC, then one can prevent reverse protection by adding a diode but for AC, I am having a hard time figuring it out.

 

For the next bit, I would draw a diode in inverse parallel to the L.E.D.

However for the last part, I am not sure how you would get the answer. Is it just half the original Resistance?

There are two ways of doing this. "inverse parallel" is fine, and it approximately balances the AC current for both polarities, though it does waste slightly more power then it uses.

Or, you put the diode same direction and in series. You need to reduce the 7V the resistor sees by the diode drop (about .7V).

For both methods for AC, since the LED only sees the AC voltage for half the time (each half cycle) then you need to double the current, so yes, the resistor halves.

One question is that why a LED needs reverse protection when there is a resistor. The Led will light in either polarity of the AC.

LEDs light only when in forward conduction. Also, they are poor diodes in that they tend to break down for a low voltage, 5 volts or so. So you need an additional diode to protect for larger voltages.

 

For a) Rmax ≈ (9-2)/8mA = 875 Ω
For b) you must put a diode with the anode to the line and the cathode to the LEd side and if Vf (diode) = 1V, Rmax ≈ (9-2-1)/8mA = 750 Ω


I hope that it helps you
greetings