(a) The function \(f(x) = 2 - \frac{5}{x+2}\) has a horizontal asymptote at \(y = 2\). As \(x \to \infty\), \(f(x) \to 2\). Therefore, the range of \(f\) is \(y < 2\).

(b) To find the inverse, start with \(y = 2 - \frac{5}{x+2}\).

Rearrange to solve for \(x\):

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

\(yx + 2y = 2x + 4 - 5\)

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

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

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

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

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

The domain of \(f^{-1}\) is determined by the range of \(f\), which is \(x < 2\).

First, complete the square for \(g(x) = x^2 - 6x + 7\).

Rewrite \(x^2 - 6x + 7\) as \((x-3)^2 - 9 + 7\).

This simplifies to \((x-3)^2 - 2\).

Set \(y = (x-3)^2 - 2\) and solve for \(x\).

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

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

Since \(x > 4\), choose the positive root: \(x = 3 + \sqrt{y + 2}\).

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

The domain of \(g^{-1}\) is \(x > -1\) because \(x + 2 > 0\).

(i) To express \(9x^2 - 6x + 6\) in the form \((ax + b)^2 + c\), complete the square:

\(9x^2 - 6x + 6 = 9(x^2 - \frac{2}{3}x) + 6\)

\(= 9((x - \frac{1}{3})^2 - \frac{1}{9}) + 6\)

\(= 9(x - \frac{1}{3})^2 - 1 + 6\)

\(= 9(x - \frac{1}{3})^2 + 5\)

Thus, \(a = 3\), \(b = -1\), \(c = 5\).

(ii) The function \(f(x) = 9x^2 - 6x + 6\) is a quadratic function that opens upwards. For \(f\) to be one-one, it must be restricted to the domain where it is either increasing or decreasing. The vertex of the parabola is at \(x = \frac{1}{3}\). Therefore, the smallest value of \(p\) is \(\frac{1}{3}\).

(iii) For \(p = \frac{1}{3}\), the function is one-one for \(x \geq \frac{1}{3}\). To find \(f^{-1}(x)\), start with \(y = (3x-1)^2 + 5\):

\(y - 5 = (3x-1)^2\)

\(\sqrt{y-5} = \pm (3x-1)\)

Since \(x \geq \frac{1}{3}\), we take the positive root:

\(3x - 1 = \sqrt{y-5}\)

\(3x = \sqrt{y-5} + 1\)

\(x = \frac{1}{3}\sqrt{y-5} + \frac{1}{3}\)

Thus, \(f^{-1}(x) = \frac{1}{3}\sqrt{x-5} + \frac{1}{3}\) with domain \(x \geq 5\).

(iv) The equation \(f(x) = q\) has no solution when \(q < 5\) because the minimum value of \(f(x)\) is 5, which occurs at the vertex \(x = \frac{1}{3}\).

(i) To express \(-x^2 + 6x - 5\) in the form \(a(x + b)^2 + c\), we complete the square:

\(-x^2 + 6x - 5 = -1(x^2 - 6x) - 5\)

Complete the square for \(x^2 - 6x\):

\(x^2 - 6x = (x - 3)^2 - 9\)

Thus, \(-1(x^2 - 6x) = -1((x - 3)^2 - 9) = -1(x - 3)^2 + 9\)

Therefore, \(-x^2 + 6x - 5 = -1(x - 3)^2 + 9 - 5 = -1(x - 3)^2 + 4\)

(ii) The function \(f(x) = -x^2 + 6x - 5\) is a downward-opening parabola. It is one-one when restricted to the domain where \(x\) is greater than or equal to the vertex. The vertex is at \(x = 3\), so the smallest \(m\) is 3.

(iii) For \(m = 5\), we find the inverse:

Start with \(y = -x^2 + 6x - 5\)

Rearrange to solve for \(x\):

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

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

\(x - 3 = \pm \sqrt{4 - y}\)

Since \(x \geq 3\), we take the positive root:

\(x = 3 + \sqrt{4 - y}\)

Thus, \(f^{-1}(x) = 3 + \sqrt{4 - x}\)

The domain of \(f^{-1}\) is \(x \leq 0\) because the range of \(f\) for \(x \geq 5\) is \(-\infty \leq y \leq 0\).

Let \(y = \frac{15}{2x+3}\).

Rearrange to solve for \(x\):

\(2x + 3 = \frac{15}{y}\)

\(2x = \frac{15}{y} - 3\)

\(x = \frac{15 - 3y}{2y}\)

Thus, \(f^{-1}(x) = \frac{15 - 3x}{2x}\).

For the range of \(f^{-1}\), since \(0 \leq x \leq 6\) for \(f(x)\), the range of \(f^{-1}\) is \(0 \leq f^{-1}(x) \leq 6\).

For the domain of \(f^{-1}\), solve \(0 \leq \frac{15}{2x+3} \leq 6\) for \(x\):

\(1 \leq x \leq 5\).