CubeyThePenguin

\(g(x) = (-2+x) - (x^3 - 2x - 2) = \boxed{-x^3+3x}\)

THe first geometric series has first term 12 and common ratio 1/3, so we can find the sum.

\(S = \frac{a}{1-r} = \frac{12}{1-\frac{1}{3}} = 18\)              .   (note that this formula only works if the absolute value of the ratio is less than 1)

For the second series, we know that the first term is 12 and the common ratio is (4+n)/12. The sum is also 4(18) = 72, so we write:

\(72 = \frac{12}{1 - \frac{4+n}{12}}\)

Solving for n, we get \(\boxed{n=6}\)

125 isn't too big of a number, so we can just use brute force. (If anyone else has a better solution, please attach it below.)

0, 1, 1, 3, 6, 9, 27, 31, 124, 129

The first term greater than 125 is 129

The first vertical column can be filled from the bottom: 18, 14, 10, 6.

In the horizontal row, we have 21, ___, ___, 6. The common difference of this is 5, so we can fill that in: 21, 16, 11, 6, 1, -4, -9.

Finally, we have the vertical column to the right. From the bottom, the numbers are -17, __, __, __, -9, N. The common difference is 2, so N = -7

Equality is achieved when \(x^2 = y^2 = \frac{1}{2}\).

\(\max({|x|+|y|}) = \frac{\sqrt{2}}{2} \cdot 2 = \boxed{\sqrt{2}}\)