(i) To express \(3x^2 - 6x + 2\) in the form \(a(x+b)^2 + c\), we complete the square:

Start with \(3(x^2 - 2x) + 2\).

Complete the square for \(x^2 - 2x\):

\(x^2 - 2x = (x-1)^2 - 1\).

Substitute back: \(3((x-1)^2 - 1) + 2 = 3(x-1)^2 - 3 + 2 = 3(x-1)^2 - 1\).

Thus, \(a = 3\), \(b = -1\), \(c = -1\).

(ii) Differentiate \(f(x) = x^3 - 3x^2 + 7x - 8\):

\(f'(x) = 3x^2 - 6x + 7\).

Complete the square for \(3x^2 - 6x + 7\):

\(3(x^2 - 2x) + 7 = 3((x-1)^2 - 1) + 7 = 3(x-1)^2 + 4\).

Since \(3(x-1)^2 + 4 > 0\) for all \(x\), \(f'(x) > 0\) for all \(x\).

Therefore, \(f\) is an increasing function.