Answer
See answers.
Work Step by Step
a. There is a voltage drop across the lines because of the resistance.
$$V_{out}=V_{in}-IR=42000V-(740A)(2)(0.95\Omega)=40594V\approx 41kV$$
b. Calculate the power input from P=IV.
$$P_{in}=IV_{in}=(740A)(42000V)=3.1\times10^7W$$
c. The power loss is due to the current and the resistance in the wires.
$$P_{loss}=I^2R=(740A)^2(2)(0.95\Omega)=1.0\times10^6W$$
d. The power output found from P=IV.
$$P_{out}=IV_{out}=(740A)(40594V)=3.0\times10^7W$$
As a check, this also equals the input power minus the loss.