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

\(gf(x) = g(f(x)) = g\left(1 + \frac{3}{x-2}\right)\).

Substitute into \(g(x) = 2x - 2\):

\(gf(x) = 2\left(1 + \frac{3}{x-2}\right) - 2\).

Simplify the expression:

\(gf(x) = 2 + \frac{6}{x-2} - 2\).

Combine like terms:

\(gf(x) = \frac{6}{x-2}\).

\((a) To solve the equation f(x) = 0, we have:\)

\(x - 4x^{\frac{1}{2}} + 1 = 0\)

\(Let y = x^{\frac{1}{2}}, then x = y^2. Substitute to get:\)

\(y^2 - 4y + 1 = 0\)

\(Using the quadratic formula, y = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\)

\(Thus, x = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}\)

(b) Given f(x) \equiv gh(x), we have:

\(gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n\)

Expanding, we get:

\(gh(x) = m(x - 4x^{\frac{1}{2}} + 4) + n\)

\(Since f(x) = x - 4x^{\frac{1}{2}} + 1, equate the expressions:\)

\(m(x - 4x^{\frac{1}{2}} + 4) + n = x - 4x^{\frac{1}{2}} + 1\)

\(Comparing coefficients, m = 1 and n = -3.\)

(b) Start with \(y = \frac{-x}{\sqrt{4-x^2}}\). Square both sides to get \(x^2 = y^2(4-x^2)\).

Rearrange to \(x^2(1+y^2) = 4y^2\).

Solving for \(x\), we get \(x = \frac{\pm 2y}{\sqrt{1+y^2}}\).

Select the correct square root: \(f^{-1}(x) = \frac{-2x}{\sqrt{1+x^2}}\).

(c) The maximum possible value of \(a\) is 1, as \(-1 < x < 1\) is required for \(fg\) to be formed.

(d) Substitute \(g(x) = 2x\) into \(f(x)\): \(fg(x) = f(2x) = \frac{-2x}{\sqrt{4-(2x)^2}}\).

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

(a) To find \(ff(5)\), we first calculate \(f(5)\):

\(f(5) = \frac{5+3}{5-1} = \frac{8}{4} = 2\).

Now, calculate \(f(f(5)) = f(2)\):

\(f(2) = \frac{2+3}{2-1} = \frac{5}{1} = 5\).

Thus, \(ff(5) = 5\).

(b) To find \(f^{-1}(x)\), set \(y = f(x) = \frac{x+3}{x-1}\).

Swap \(x\) and \(y\):

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

Cross-multiply to solve for \(y\):

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

\(xy - x = y + 3\).

\(xy - y = x + 3\).

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

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

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

(a) We have \(fg(x) = f(g(x)) = (g(x))^2 - 1\). Given \(fg(x) = 3\), we set \((g(x))^2 - 1 = 3\), leading to \((g(x))^2 = 4\).

Since \(g(x) = \frac{1}{2x+1}\), we have \(\left(\frac{1}{2x+1}\right)^2 = 4\).

This implies \(\frac{1}{(2x+1)^2} = 4\), leading to \((2x+1)^2 = \frac{1}{4}\).

Solving \(2x+1 = \pm \frac{1}{2}\), we get \(2x+1 = \frac{1}{2}\) or \(2x+1 = -\frac{1}{2}\).

For \(2x+1 = -\frac{1}{2}\), solving gives \(x = -\frac{3}{4}\).

For \(2x+1 = \frac{1}{2}\), solving gives \(x = -\frac{1}{4}\), but this does not satisfy \(x < -\frac{1}{2}\).

Thus, \(x = -\frac{3}{4}\) only.

(b) Let \(y = fg(x) = (g(x))^2 - 1\).

Then \(y = \frac{1}{(2x+1)^2} - 1\), leading to \(\frac{1}{(2x+1)^2} = y + 1\).

Thus, \((2x+1)^2 = \frac{1}{y+1}\).

Taking square roots, \(2x+1 = \pm \frac{1}{\sqrt{y+1}}\).

Solving for \(x\), we get \(x = -\frac{1}{2\sqrt{y+1}} - \frac{1}{2}\).