Electricity: A 440/110 V single phase transformer supplies a load of 5 kW at 0.8 power factor...

5kva is the load, not the supply.

IDk if this is right but this is how I did it.
5,000W / 0.8 = 6250W

6250W /110V = 56.82A

P.F = 0.8 is equal to both sides of the xfrmr

110(.8) = secondary voltage

P=IE

5000w = 88I

56.82= I

Hubert page 184-185

AC Power calculation P=EI*PF

P=load power
E=secondary voltage where load is connected
PF= power factor

solve for I, I=P/(E*PF)= 5000/(110*0.8)= 56.818A