(ii) To express \(f(x) = x^2 - 3x\) in the form \((x-a)^2 - b\), complete the square:

\(x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4}\).

Thus, \(a = \frac{3}{2}\) and \(b = \frac{9}{4}\).

(iii) The minimum value of \(f(x) = (x - \frac{3}{2})^2 - \frac{9}{4}\) is \(-\frac{9}{4}\), so the range is \(f(x) \geq -\frac{9}{4}\).

(iv) The function \(f(x) = x^2 - 3x\) is not one-to-one because it is a quadratic function, which is symmetric about its vertex. Therefore, \(f\) does not have an inverse.

(v) Solve \(g(x) = x - 3\sqrt{x} = 10\).

Let \(y = \sqrt{x}\), then \(x = y^2\).

Substitute into the equation: \(y^2 - 3y = 10\).

Rearrange to form a quadratic: \(y^2 - 3y - 10 = 0\).

Factorize: \((y - 5)(y + 2) = 0\).

Thus, \(y = 5\) or \(y = -2\).

Since \(y = \sqrt{x} \geq 0\), \(y = 5\).

Therefore, \(x = y^2 = 25\).