(i) The function \(f(x) = x^2 + 4x\) can be rewritten as \((x+2)^2 - 4\). The vertex of this parabola is at \(x = -2\), giving the minimum value \(-4\). For \(f\) to be one-one, \(x \geq c\) must be such that the function is increasing, which occurs for \(x \geq -2\). Thus, the range is \(y \geq c^2 + 4c\) and the smallest possible value of \(c\) is \(-2\).

(ii) Given \(gf(1) = 11\) and \(fg(1) = 21\), we have:

1. \(g(f(1)) = g(5) = 5a + b = 11\)

2. \(f(g(1)) = f(a + b) = (a + b)^2 + 4(a + b) = 21\)

Substituting \(b = 11 - 5a\) into the second equation:

\((a + (11 - 5a))^2 + 4(a + (11 - 5a)) = 21\)

\((11 - 4a)^2 + 4(11 - 4a) = 21\)

\(121 - 88a + 16a^2 + 44 - 16a = 21\)

\(16a^2 - 104a + 144 = 21\)

\(16a^2 - 104a + 123 = 0\)

Solving this quadratic equation gives \(a = 2\) and \(b = 1\).