Let \(y = \frac{5}{1 - 3x}\).

Rearrange to make \(x\) the subject:

\(3x = 1 - \frac{5}{y}\)

\(3x = \frac{y - 5}{y}\)

\(x = \frac{y - 5}{3y}\)

Thus, \(f^{-1}(x) = \frac{x - 5}{3x}\) or \(\frac{1}{3} - \frac{5}{3x}\).

The range of \(f\) is \(x \geq 1\), so the domain of \(f^{-1}\) is \(-2.5 \leq x < 0\).

The domain of \(f\) is \(x \geq 1\), so the range of \(f^{-1}\) is \(x \geq 1\).

(i) To express \(2x^2 - 12x + 13\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 12x + 13 = 2(x^2 - 6x) + 13\)

\(= 2((x-3)^2 - 9) + 13\)

\(= 2(x-3)^2 - 18 + 13\)

\(= 2(x-3)^2 - 5\)

Thus, \(a = 2, b = -3, c = -5\).

(ii) For \(f(x)\) to be one-one, the derivative \(f'(x) = 4x - 12\) must be non-negative for \(x \geq k\). Solving \(4x - 12 \geq 0\) gives \(x \geq 3\). Therefore, the smallest possible value of \(k\) is 3.

(iii) With \(k = 7\), the minimum value of \(f(x)\) is \(f(7)\):

\(f(7) = 2(7)^2 - 12(7) + 13 = 98 - 84 + 13 = 27\)

Thus, the range of \(f\) is \(y \geq 27\).

(iv) To find \(f^{-1}(x)\), start with \(y = 2(x-3)^2 - 5\) and solve for \(x\):

\(2(x-3)^2 = y + 5\)

\((x-3)^2 = \frac{1}{2}(y + 5)\)

\(x-3 = \pm \sqrt{\frac{1}{2}(y + 5)}\)

Since \(x \geq 7\), choose the positive root:

\(x = 3 + \sqrt{\frac{1}{2}(y + 5)}\)

Thus, \(f^{-1}(x) = 3 + \sqrt{\frac{1}{2}(x + 5)}\) for \(x \geq 27\).