(1)如图所示,消耗的电能:W=825.4kW?h-675.2kW?h=150.2kW?h;(3)电能表转400次消耗的电能:W′= 400 3200 kW?h=0.125kW?h,则电水壶的实际功率:P′= W′ t = 0.125kW?h 10 60 h =0.75kW=750W.故答案为:150.2;750.