(i) To solve \(gf(x) = x\), substitute \(f(x) = 2x + 1\) into \(g(x)\):

\(gf(x) = \frac{2(2x + 1) - 1}{2x + 1 + 3} = \frac{4x + 1}{2x + 4}\)

Set \(\frac{4x + 1}{2x + 4} = x\) and solve for \(x\):

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

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

\(2x^2 = 1\)

\(x^2 = \frac{1}{2}\)

\(x = \frac{1}{2}\sqrt{2}\)

(ii) To find \(f^{-1}(x)\), solve \(y = 2x + 1\) for \(x\):

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

Thus, \(f^{-1}(x) = \frac{1}{2}(x - 1)\).

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

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

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

\(yx - 2x = -3y - 1\)

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

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

Thus, \(g^{-1}(x) = \frac{-1 - 3x}{x - 2}\) or \(\frac{1 + 3x}{2 - x}\).

(iii) To show \(g^{-1}(x) = x\) has no solutions, set \(\frac{-1 - 3x}{x - 2} = x\):

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

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

\(x^2 + x + 1 = 0\)

The discriminant \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\), which is negative, indicating no real roots.