(a) To find \(fg(x)\), substitute \(g(x) = 2x + 1\) into \(f(x)\):

\(fg(x) = f(g(x)) = (2x+1)^2 + 3\).

(b) Let \(y = (2x+1)^2 + 3\). Solve for \(x\):

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

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

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

Since \(x > 0\), choose the positive root:

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

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

(c) Solve \(fg(x) - 3 = gf(x)\):

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

\((2x+1)^2 = 2x^2 + 6 + 1\)

\(4x^2 + 4x + 1 = 2x^2 + 7\)

\(2x^2 + 4x - 6 = 0\)

\(2(x+3)(x-1) = 0\)

\(x = 1\)