(i) To express \(2x^2 + 8x - 10\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 + 8x - 10 = 2(x^2 + 4x) - 10\)

\(= 2((x+2)^2 - 4) - 10\)

\(= 2(x+2)^2 - 8 - 10\)

\(= 2(x+2)^2 - 18\)

Thus, \(a = 2\), \(b = 2\), \(c = -18\).

(ii) The least value of \(y\) occurs at the vertex of the parabola, which is at \(x = -2\). Substituting \(x = -2\) into the equation:

\(y = 2(-2)^2 + 8(-2) - 10 = 8 - 16 - 10 = -18\)

Thus, the least value of \(y\) is \(-18\) when \(x = -2\).

(iii) To find the set of values of \(x\) for which \(y \geq 14\), solve:

\(2x^2 + 8x - 10 \geq 14\)

\(2x^2 + 8x - 24 \geq 0\)

\(x^2 + 4x - 12 \geq 0\)

\((x+6)(x-2) \geq 0\)

The solution is \(x \geq 2\) or \(x \leq -6\).

(iv) For \(f\) to be one-one, the domain must be restricted to where the function is either increasing or decreasing. The vertex is at \(x = -2\), so the smallest \(k\) is \(-2\).

(v) To find \(f^{-1}(x)\), solve \(y = 2x^2 + 8x - 10\) for \(x\):

\(y = 2(x+2)^2 - 18\)

\(y + 18 = 2(x+2)^2\)

\(\frac{y+18}{2} = (x+2)^2\)

\(x+2 = \pm \sqrt{\frac{y+18}{2}}\)

Since \(x \geq -2\), choose the positive root:

\(x = \sqrt{\frac{y+18}{2}} - 2\)

Thus, \(f^{-1}(x) = \sqrt{\frac{x+18}{2}} - 2\).