What are my power requirements for my second resistor, and did I do my first's power calculation correctly for my first LED floodlight design?
Your math is wrong. The wattage wasted on R1 is P = (VSource - VLed) * I. In this case, its ~25 Watts, not 36 Watts. That's what R1's wattage should be calculated for.

To calculate R2, you would just add R2 to R1 as series resistors, then taking IResistor = (VSource - VLed) / (R1 + R2). Then IResistor * R1 for the voltage across R1, IResistor * R2 for the voltage across R2, then multiple that voltage with the current to find power.

But that's hella wasteful. 29% Efficiency. A switching constant current led driver, or using 3 3W Leds in series instead of a single 10W would be much much better. Or 3 10W in series! Your wasting 70 Percent of your battery power in the resistor heat.