So here are the solutions to the 13 coin conundrum .
Part a was solved in the comments but to rehash it.
You begin by observing that the 13 coins must all be the same weight mod 2. If this was false, i.e. if there was a coin of even weight and a coin of odd weight then removing one of them would leave the rest of the coins with a total weight which is odd, and hence that cannot be split into 2 equal bits.
One then iterates this observation by observing that they must be the same weight mod 4,8,16 etc. Eventually you get beyond the size of the largest coin and hence they are the same weight. The easiest way to see this is to divide the weights by 2 if they are all even and to ad 1 and divide by 2 if they are all odd.
Part b requires less ingenuity but more advanced techniques. Consider the vector space over Z generated by the 13 weights. It is clearly finite dimensional. Write each weight as a vector in this space and apply the result of part a to each co-ordinate.
Similiarly for part c it's a vector space of size 2 over R (i.e. part b).