Task 1, CJam, 7 bytes
I just wanted to get the (presumably) optimal CJam solution for this in. It makes use of the fact that the sum of the first n cubes is the square of the nth triangular number, which is itself the sum of the first n integers.
Task 9, Mathematica, 4 bytes
This is essentially the same as Sp3000's solution, but using more syntactical trickery (and since I'm losing the tie to him, I don't think there's any harm in that). The
## is a sequence with all parameters of the unnamed function. Now
1## is short for
1*##, which is syntactical sugar for
## gets expanded into the number and the nested list. Multiplication threads over lists automatically in Mathematica.