(i) To find \(f^{-1}(x)\), start with \(y = 2x + 5\). Solve for \(x\):

\(y - 5 = 2x\)

\(x = \frac{1}{2}(y - 5)\)

Thus, \(f^{-1}(x) = \frac{1}{2}(x - 5)\).

For \(g^{-1}(x)\), start with \(y = \frac{8}{x-3}\). Solve for \(x\):

\(y(x - 3) = 8\)

\(yx - 3y = 8\)

\(yx = 8 + 3y\)

\(x = \frac{8}{y} + 3\)

Thus, \(g^{-1}(x) = \frac{8}{x} + 3\), and it is not defined for \(x = 0\).

(ii) Given \(fg(x) = 5 - kx\), we have:

\(fg(x) = f\left(\frac{8}{x-3}\right) = 2\left(\frac{8}{x-3}\right) + 5 = \frac{16}{x-3} + 5\)

Set \(\frac{16}{x-3} + 5 = 5 - kx\)

\(\frac{16}{x-3} = -kx\)

\(16 = -kx(x-3)\)

\(kx^2 - 3kx + 16 = 0\)

For no solutions, the discriminant must be less than zero:

\(b^2 - 4ac = (-3k)^2 - 4(k)(16) < 0\)

\(9k^2 - 64k < 0\)

\(k(9k - 64) < 0\)

The critical points are \(k = 0\) and \(k = \frac{64}{9}\).

Thus, the set of values is \(0 < k < \frac{64}{9}\).