a.) How much work is required of the pump to deliver 2500J of heat into the house if the outdoor temperature is 0 degrees. Assume an ideal (Carnot) coefficient of performance COP= T_H/(T_HT_L).
b.) How much work is required of the pump to deliver 2500J of heat into the house if the outdoor temperature is 15 degrees. Assume an ideal (Carnot) coefficient of performance COP= T_H/(T_HT_L).
2 Answers

For both cases, Work = Energy / COP. You stated the formula for COP that you can calculate, so why can not you do it?
Perhaps, I can see your problem now, in not converting to the temperatures to Kelvin before hand. Add 273 to each temperature reading.

The heating load is the heat to be rejected into the house being heater, QR.
The Heat Pump Heater COP, is defined by :
COP = QR / WHP
where WHP is the required Heat Pump Heater work.
For a Carnot Heat Pump Heater, you have :
COP = ( TH ) / ( TH  TL )
For part a you have :
QR = 2500 J
COP = ( TH ) / ( TH  TL ) = ( 295.2 K ) / ( 295.2 K  273.2 K ) = 13.42 COP Units
WHP = QR / COP = 2500 J / 13.42 = 186 J <[ a ]
For Part b, you have :
COP = ( TH ) / (TH  TL ) = ( 295.2 K ) / ( 295.2 K  258.2 K ) = 7.978 COP Units
WHP = QR / COP = ( 2500 J ) / ( 7.978 COP units ) = 313 J <[ b ]