Impedance across AC circuit
I assume:

"source amplitude" means "peak" (as apposed to peak to peak)

The load impedance is actually calculated at 60Hz

Ztotal = 1000 + 1000j  (really should be using j as sqrt(-1) not i in electronics)

Ztotal = 1414.21 @ 45deg (just rewritten in a magnitude / phase representation)

|Ztotal| = 1414.21 Ohms

Vrms = 10/sqrt(2) ~= 7.07 Vrms

|I| = |Vrms| / |Ztotal| = 7.07 / 1414.21

|I| ~= 5mA

P(true power) = I^2 * R = (5mA)^2 * 1000 = 25 mW (this is what is dissipated in the load, your answer)

Q(reactive power) = I^2 * X = (5mA)^2 * 1000 = 25 mVAR (this is power bouncing back and forth, not dissipated)

S(apparent power) = I^2 * Z = (5mA)^2 * 1414.21 = 35.35 mVA (this is the vector sum of the true and reactive power)

Power Factor = P/S = True Power / Apparent Power = 25mW / 35mVA = 0.714

Hopefully I didn't flub up the math there, my calculators batteries just died so i did it all in google.