(a) To express \(f(x) = x^2 - 4x + 3\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x^2 - 4x + 4) - 4 + 3 = (x-2)^2 - 1\).

(b) Since \(f\) is a one-one function, the smallest possible value of \(c\) is the vertex of the parabola, which is \(x = 2\).

(c) Given \(c = 5\), the function \(f(x) = (x-2)^2 - 1\) is one-one for \(x > 5\). To find \(f^{-1}(x)\), set \(y = (x-2)^2 - 1\):

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

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

Since \(x > 5\), choose the positive root: \(x = 2 + \sqrt{y+1}\).

Thus, \(f^{-1}(x) = 2 + \sqrt{x+1}\) for \(x > 8\).

(d) To find \(gf(x)\), substitute \(f(x)\) into \(g(x)\):

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

The range of \(gf(x)\) is \(0 < gf(x) < \frac{1}{9}\) since \((x-2)^2 > 0\) and the minimum value of \((x-2)^2\) is 9 when \(x = 5\).