New to EE classes, have an introductory voltage output problem
Assuming that no power is actually supplied to the load, your first task is to calculate the dropping resistors. Use $$P =\frac{V^2}{R} $$ where V is 24. 

Now divide that resistance in half, and use 2 such resistors as per your schematic.

Finally, divide your R2 into 2 resistors set up as a voltage divider, with the sum of the resistors equal to the R2 value, with 12 volts divided down to 5 volts. (Hint - the ratio of the two will be 5:7, right?) 

EDIT - Since you have made it clear that the 80 mW figure applies "to the load", it is also clear from your schematic that the 80 mW figure must apply to the 5 volt supply only, since as long as R1 and R2 are identical the supply voltages will be equal to +/- 12 (assuming equal loads). So, since $$P = iV $$ $$i=\frac{P}{V} =\frac{.08}{5} = .016 = 16\text{ mA}$$Then the 5 volt load can be modelled as 312.5 ohn resistor, and you can figure a voltage divider which will produce that. Note that this resistor network will need to be completely separate from your R1/R2 network, and the reference points cannot be common. In other words, there will not be a common ground point for +5 and +/-12. The reason is that if you do so, the load resistances (which are not specified, and so must be assumed to be arbitrary) will interact with the 5 volt network and change the 5 volt level.

All in all, this is a very confusing, poorly defined problem, and unless you have failed to pass on information which was contained in the problem as it was given to you, I suggest that you are dealing with a bad course.