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

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

(ii) The function \(f(x) = x^2 - 4x + 7\) is decreasing when its derivative \(f'(x) = 2x - 4\) is negative. Solving \(2x - 4 < 0\) gives \(x < 2\). Therefore, the largest value of \(k\) is 2, so \(k \leq 2\).

(iii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\). Solving for \(x\), we get:

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

Since \(x < 1\), choose the negative root: \(x = 2 - \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 - \sqrt{x-3}\) for \(x > 4\).

(iv) The composition \(gf(x)\) is given by:

\(gf(x) = g(f(x)) = \frac{2}{x^2 - 4x + 7 - 1} = \frac{2}{(x-2)^2 + 2}\).

The range of \(gf(x)\) is determined by the values of \((x-2)^2 + 2\), which is always greater than 2. Therefore, the range of \(gf(x)\) is \(0 < gf(x) < \frac{2}{3}\).