(i) Completing the square for \(f(x) = x^2 + 6x - 8\), we have:

\(x^2 + 6x - 8 = (x+3)^2 - 17\)

The minimum value of \((x+3)^2\) is 0, so the minimum value of \(f(x)\) is \(-17\). Thus, the range of \(f(x)\) is \(\geq -17\).

(ii) Given roots \(x = k\) and \(x = -2k\), the equation can be written as:

\((x-k)(x+2k) = 0\)

Expanding gives \(x^2 + 5kx - 2k^2 = 0\). Comparing with \(x^2 + 5x + b = 0\), we find:

\(5k = 5 \Rightarrow k = 5\)

\(b = -2k^2 = -2(5)^2 = -50\)

(iii) For \(f(x+a) = a\), substitute \(x+a\) into \(f\):

\((x+a)^2 + a(x+a) + b = a\)

\(x^2 + 2ax + a^2 + ax + a^2 + b = a\)

\(x^2 + (2a+a)x + (a^2 + b - a) = 0\)

Using the discriminant \(b^2 - 4ac\) for no real roots:

\((3a)^2 - 4(1)(a^2 + b - a) < 0\)

\(9a^2 - 4(a^2 + b - a) < 0\)

\(9a^2 - 4a^2 - 4b + 4a < 0\)

\(5a^2 < 4(b-a)\)

\(a^2 < 4(b-a)\)