The effective resistance seen by the voltage source is 5.4 ohm.
Hint for 2.11
P = I.V
kilowatt-hour (kW-hr) is the conventional unit for electrical energy.
Hint for 2.13
The terminal voltage of the non-ideal source is the same as the voltage across the load. The internal resistance associated with the source drops the terminal voltage below the emf whenever current is flowing.
Hint for 2.38
As mentioned on the errata page, I believe that the load should be characterized as 5 ohm, rather than 5 W. Although loads are often characterized by the power they dissipate when connected to a 115 ac source, it is unusual to see it characterized this way for an arbitrary voltage. Also, notice that part b) asks what happens when the load is changed to 10 ohm.
Hint for 2.40
So how much resistance can be tolerated if we wish to drop a voltage of only 5 V at a current of 7.1 A?
I do not find an adequate AWG guide in our textbook. The first one I found using a web search is here.
Hint for 2.54
Use the terminal voltage across 10 ohm to find the current flowing through the battery. Then use the difference between the emf and the terminal voltage to find the internal resistance responsible for the voltage drop at this current.