In the second part of the question, setting N = 100a + 10b + c and S = a + b^2 + c^3 and setting N = S, I can then form this equation:
99a + b (b-1) = c (c+1) (c-1), drawing on results from the first part of the Q, in doing so.
To make my work a bit clearer I I defined the following terms:
x = 99a, y = b (b-1), z = c (c+1) (c-1)
As there are only a small, limited amount of integers equal to c(c+1)(c-1) between 100 and 1000 we can use them to figure out x and y.
There are also a limited amount of numbers that are multiples of 99 between 100 and 1000.
y must be also below 25 so this means z-x < 25.
Fortunately there are only a few permutations of x and z that satisfy this. 3 in total. However one is omitted as it would result in b being a non-integer solution. As y is a quadratic, altogether we would have 4 solutions of b meaning 4 solutions to N=S.