Part A: Design and construct a circuit that puts out 8v (within 0.2v) from a 12v power supply, with no load attached to the two output terminals. THEN..
Add: A the equivalent of a 1kΩ load so that the output voltage drops to 4v (within 0.2v)

 

Isn't the "added" load resistor supposed to be 1KΩ (not 680Ω)?

 

I think you are misunderstanding the point of the question.

You first design a circuit that cuts 12 V down to 8 V when there is no load. There are many, many choices of resistors that you could make to accomplish this.

But now, given your choice of resistors, you add a load that has an equivalent resistance of 1 kΩ (the actual load might be very complicated, but as far as your circuit is concerned it behaves as if it were just a 1 kΩ resistor) across the (unloaded) 8 V output terminals. Now what is the output voltage? That will depend on the choice of resistors you made in the first part. So you need to revisit that choice and redo it in order to get that loaded output voltage to be 4 V.

If you plan ahead, you can come up with a pair of equations that, given a desired open-circuit output and a desired loaded output, will give you the ideal choice for each of the two resistors.

Are you constrained to only using standard resistor values? If not, then I don't understand why the 0.2 V tolerance since you can hit it exactly. If so, then bear in mind that neither 500 Ω nor 320 Ω are standard values.