(i) To find the set of values of \(x\) for which \(f(x) > 15\), solve the inequality:

\(x^2 - 2x > 15\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 15 = 0\)

Factorize the quadratic:

\((x - 5)(x + 3) = 0\)

The solutions are \(x = 5\) and \(x = -3\).

Test intervals around the roots to find where \(f(x) > 15\):

For \(x < -3\), choose \(x = -4\):

\(f(-4) = 16 + 8 = 24 > 15\)

For \(x > 5\), choose \(x = 6\):

\(f(6) = 36 - 12 = 24 > 15\)

Thus, \(x < -3\) and \(x > 5\).

(ii) To find the range of \(f\), complete the square:

\(f(x) = x^2 - 2x = (x-1)^2 - 1\)

The minimum value of \((x-1)^2\) is 0, occurring at \(x = 1\).

Thus, the minimum value of \(f(x)\) is \(-1\).

Therefore, the range of \(f\) is \(y \geq -1\).

Since \(f\) is a quadratic function and not one-to-one, it does not have an inverse.