(i) To express \(4x^2 + 12x + 10\) in the form \((ax + b)^2 + c\), we complete the square:

\(4x^2 + 12x + 10 = 4(x^2 + 3x) + 10\).

Complete the square for \(x^2 + 3x\):

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

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

Therefore, \(4x^2 + 12x + 10 = 4(x + \frac{3}{2})^2 - 9 + 10 = 4(x + \frac{3}{2})^2 + 1\).

So, \((ax + b)^2 + c = (2x + 3)^2 + 1\) with \(a = 2, b = 3, c = 1\).

(ii) Given \(fg(x) = 4x^2 + 12x + 10\) and \(f(x) = x^2 + 1\), we have:

\(g(x) = \frac{4x^2 + 12x + 10}{x^2 + 1}\).

From part (i), \(4x^2 + 12x + 10 = (2x + 3)^2 + 1\), so:

\(g(x) = \frac{(2x + 3)^2 + 1}{x^2 + 1} = 2x + 3\).

(iii) To find \((fg)^{-1}(x)\), solve \(y = (2x + 3)^2 + 1\) for \(x\):

\(y - 1 = (2x + 3)^2\).

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

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

Choose the positive root for \(x > 0\):

\((fg)^{-1}(x) = \frac{1}{2}\sqrt{x-1} - \frac{3}{2}\).

The domain of \((fg)^{-1}\) is \(x > 10\) since \(y = (2x + 3)^2 + 1 > 10\).