1. The problem statement, all variables and given/known data I don't know where to start. Can anyone recommend a book the instructor didn't assign one.
I don't think a book is necessary, as long as you know how to find the equivalent resistances for parallel and series circuits. For the first question, imagine a current I entering at A and leaving through B. How much current will flow through each of the resistors? You can use this to figure out the voltage drop between the two points. For the second question, suppose I take off the 3 resistors at the very right. What would the equivalent resistance be between the top-right point and the bottom-right point? You can use this to set up an equation to solve for the equivalent resistance.
Welcome to PhysicsForums! These are more math / problem solving-type questions, rather than straightforward circuit analysis. Here's a hint for a: consider what happens when you inject a test current 'I_o' into point A, and figure out how the current flows. Then figure out the voltage at each point (V=IR) For b, consider that you're adding three series resistors in parallel with a single one. And then you're adding another three series resistors in parallel with the one resistor that you just added previously. Consider: you're far down the line, does adding another three resistors change the equivalent resistance by very much? You should end up solving a quadratic equation using this approach. Good luck! EDIT: Obviously, adding three resistors at that point changes the resistance at that point. It shouldn't change the resistance very much at the beginning of your resistor chain.
The second diagram looks too easy if Points A and B on the circuit are the far right top and bottom "corners" on the circuit. There is only one resistor between points A and B? What am I missing?
The first one is done by joining together (with wires) points on the cube that must have the same voltage on them, by symmetry. If they have the same voltage, no current will flow in the new wires, but they make the problem a lot easier. You get a few sets of parallel resistors in series.