(i) Since \(f(x)\) is divisible by \(x^2 - 4x + 4\), we can write \(f(x) = (x^2 - 4x + 4)(x^2 + bx + c)\).

By comparing coefficients, we find that \(b = 2\) and \(4c = a\). Solving for \(a\), we get \(a = 8\).

(ii) Substitute \(a = 8\) into \(f(x)\), giving \(f(x) = (x^2 - 4x + 4)(x^2 + 2x + 2)\).

Since \(x^2 - 4x + 4 = (x-2)^2\), it is always non-negative.

For \(x^2 + 2x + 2\), the discriminant is \(2^2 - 4 \times 1 \times 2 = -4\), which is negative, indicating no real roots and always positive.

Thus, \(f(x)\) is the product of two non-negative expressions, making \(f(x)\) never negative.