(i) Differentiate \(f(x) = (x-2)^2 g(x)\) to get \(f'(x) = 2(x-2)g(x) + (x-2)^2 g'(x)\). Since \((x-2)\) is a common factor, \((x-2)\) is a factor of \(f'(x)\).

(ii) Given \(x^5 + ax^4 + 3x^3 + bx^2 + a\) has a factor \((x-2)^2\), substitute \(x = 2\) to get \(32 + 16a + 24 + 4b + a = 0\), simplifying to \(32 + 17a + 4b = 0\).

Differentiate the polynomial to get \(5x^4 + 4ax^3 + 9x^2 + 2bx\). Substitute \(x = 2\) to get \(80 + 32a + 36 + 4b = 0\), simplifying to \(116 + 32a + 4b = 0\).

Solving the equations \(32 + 17a + 4b = 0\) and \(116 + 32a + 4b = 0\) simultaneously, we find \(a = -4\) and \(b = 3\).