(a) The function \(f(x) = 1 + \frac{3}{x-2}\) has a vertical asymptote at \(x = 2\) and a horizontal asymptote at \(y = 1\). As \(x \to \infty\), \(f(x) \to 1\). Therefore, the range of \(f\) is \(y > 1\).

(b) To find \(f^{-1}(x)\), start with \(y = 1 + \frac{3}{x-2}\).

Rearrange to find \(x\):

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

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

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

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

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

The domain of \(f^{-1}\) is \(x > 1\) because the original range of \(f\) was \(y > 1\).

To find the inverse function \(f^{-1}(x)\), we start with the equation:

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

Rearrange to make \(x\) the subject:

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

Square both sides:

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

Multiply both sides by 2:

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

Expand \((y - 1)^2\):

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

\(2y^2 - 4y + 2 = x + 3\)

Rearrange to solve for \(x\):

\(x = 2y^2 - 4y + 2 - 3\)

\(x = 2y^2 - 4y - 1\)

Thus, \(f^{-1}(x) = 2x^2 - 4x - 1\).

For the domain of \(f^{-1}\), since \(f(x) = \sqrt{\frac{x+3}{2}} + 1\) is defined for \(x \geq -3\), the range of \(f(x)\) is \(y \geq 1\). Therefore, the domain of \(f^{-1}\) is \(x \geq 1\).

(i) To express \(f(x) = 4x^2 - 24x + 11\) in the form \(a(x-b)^2 + c\), complete the square:

\(f(x) = 4(x^2 - 6x) + 11\)

\(= 4((x-3)^2 - 9) + 11\)

\(= 4(x-3)^2 - 36 + 11\)

\(= 4(x-3)^2 - 25\)

The vertex form is \(4(x-3)^2 - 25\), so the vertex is \((3, -25)\).

(ii) The range of \(g(x) = 4x^2 - 24x + 11\) for \(x \leq 1\) is determined by the vertex, which is the minimum point. Thus, the range is \(g(x) \geq -9\).

(iii) To find \(g^{-1}(x)\), start with \(y = 4(x-3)^2 - 25\).

\((x-3)^2 = \frac{1}{4}(y + 25)\)

\(x-3 = \pm \frac{1}{2}\sqrt{y + 25}\)

Since \(x \leq 1\), choose the negative root:

\(x = 3 - \frac{1}{2}\sqrt{y + 25}\)

Thus, \(g^{-1}(x) = 3 - \frac{1}{2}\sqrt{x + 25}\).

The domain of \(g^{-1}\) is \(x \geq -9\) because the range of \(g\) is \(g(x) \geq -9\).

Let \(y = \frac{3}{2x+5}\).

Rearrange to make \(x\) the subject: \(2x + 5 = \frac{3}{y}\).

Multiply both sides by \(y\): \(y(2x + 5) = 3\).

Expand: \(2xy + 5y = 3\).

Rearrange to isolate \(x\): \(2xy = 3 - 5y\).

Divide by \(2y\): \(x = \frac{3 - 5y}{2y}\).

Thus, \(f^{-1}(x) = \frac{3 - 5x}{2x}\).

Let \(y = f(x) = (3x + 2)^3 - 5\).

To find \(f^{-1}(x)\), solve for \(x\) in terms of \(y\):

\(y = (3x + 2)^3 - 5\)

Add 5 to both sides: \(y + 5 = (3x + 2)^3\)

Take the cube root: \(\sqrt[3]{y + 5} = 3x + 2\)

Subtract 2: \(\sqrt[3]{y + 5} - 2 = 3x\)

Divide by 3: \(x = \frac{\sqrt[3]{y + 5} - 2}{3}\)

Thus, \(f^{-1}(x) = \frac{\sqrt[3]{x + 5} - 2}{3}\).

The domain of \(f^{-1}\) is the range of \(f\), which is \(x \geq 3\).