Double check on #2; remember that a circuit requires _two_ conductors, and you will have voltage drop in _each_. In other words, the length of wire over which you need a maximum of 3% voltage drop will be 2*2800'.

For #3 it appears that the equations compensate for the return path, by multiplying the length over which you calculate the voltage drop by 1.732 (the square root of 3).

I've never worked with k in units of ohm*foot/cmil, but when I calculated it for aluminium from k given in ohm-cm, I got 16.7 at 20C, which agrees with 17. I didn't double

-Jon