19,20,38,41,45,80,83,85

Hint: You need to evaluate in which situation will the current *I* be larger. Use P=V*I

Hint: You need to use $I=P/V$, $P=I^2/R$. In this equation you see that the larger the voltage, the smaller the current. A smaller current involves less dissipation in heat, and therefore less waste of power. Evaluate P1 and P2 for the two different currents.

Hint: You do not need to use the mass. It is only for distraction!
For part a you need to evaluate the horsepower of the car. Use P=F*V and convert the result to HP.
For part b you can do many things. What I would do is obtain the total charge of the battery.

Qtotal = number of batteries* Q of 1 battery (note that 1 A h=3600 C).

The total energy that the battery can deliver is Qtotal*V. Note that V is 12 (batteries in parallel).

Then evaluate how long will the car run with this energy using that this energy is totally used in friction energy

U$_{friction}$=F*X. And obtain *X*!

You need to use $P=V^2/R$ in the two situations.

You need to use I=Q/t and l=v*t . $l=2*\pi*R$ . From here you can obtain the total charge and then dive this total charge by the charge of 1 proton.