(i) To express \(f(x) = x^2 - 4x + 7\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x-2)^2 - 4 + 7 = (x-2)^2 + 3\).

The range of \(f\) is \(f(x) > 3\) since \(x > 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\).

Rearrange to solve for \(x\):

\(x - 2 = \pm \sqrt{y - 3}\).

Since \(x > 2\), we take the positive root:

\(x = 2 + \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 + \sqrt{x-3}\).

The domain of \(f^{-1}\) is \(x > 3\).

(iii) Given \(f = hg\) and \(g(x) = x - 2\), we have:

\(f(x) = h(x)g(x) = h(x)(x-2)\).

Since \(f(x) = x^2 - 4x + 7\), equate:

\(h(x)(x-2) = x^2 - 4x + 7\).

Assume \(h(x) = x^2 + 3\) to satisfy the equation.