(i) For \(f(x) = \frac{3}{2x+1}\), as \(x \to 0^+\), \(f(x) \to 3\). As \(x \to \infty\), \(f(x) \to 0\). Thus, the range of \(f\) is \(0 < f(x) < 3\).

For \(g(x) = \frac{1}{x} + 2\), as \(x \to 0^+\), \(g(x) \to \infty\). As \(x \to \infty\), \(g(x) \to 2\). Thus, the range of \(g\) is \(g(x) > 2\).

(ii) \(fg(x) = f(g(x)) = \frac{3}{2\left(\frac{1}{x} + 2\right) + 1} = \frac{3}{\frac{2}{x} + 4 + 1} = \frac{3}{\frac{2 + 5x}{x}} = \frac{3x}{2 + 5x}\).

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

Rearranging gives \(2y + 5xy = 3x\) or \(3x - 5xy = 2y\).

Solving for \(x\), we get \(x(3 - 5y) = 2y\) and \(x = \frac{2y}{3 - 5y}\).

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

(i) To find \(f^{-1}(x)\), start with \(y = 3x - 2\). Solve for \(x\):

\(y + 2 = 3x\)

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

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

For \(g^{-1}(x)\), start with \(y = \frac{2x + 3}{x - 1}\). Solve for \(x\):

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

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

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

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

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

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

\(g^{-1}(x)\) is not defined for \(x = 2\).

(ii) To solve \(fg(x) = \frac{7}{3}\), first find \(fg(x)\):

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

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

Set \(\frac{3(2x + 3)}{x - 1} - 2 = \frac{7}{3}\)

\(\frac{3(2x + 3)}{x - 1} = \frac{7}{3} + 2\)

\(\frac{3(2x + 3)}{x - 1} = \frac{13}{3}\)

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

\(3(2x + 3) = 13(x - 1)\)

\(6x + 9 = 13x - 13\)

\(6x - 13x = -13 - 9\)

\(-7x = -22\)

\(x = -8\)

(i) To express \(-2x^2 + 12x - 3\) in the form \(-2(x+a)^2 + b\), we complete the square:

\(-2x^2 + 12x - 3 = -2(x^2 - 6x) - 3\)

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

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

Substitute back:

\(-2((x-3)^2 - 9) - 3 = -2(x-3)^2 + 18 - 3 = -2(x-3)^2 + 15\)

Thus, \(a = -3\) and \(b = 15\).

(ii) The greatest value of \(f(x)\) is the maximum value of \(-2(x-3)^2 + 15\), which occurs when \((x-3)^2 = 0\). Therefore, the greatest value is 15.

(iii) To find \(x\) for which \(gf(x) + 1 = 0\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = 2(-2x^2 + 12x - 3) + 5 = -4x^2 + 24x - 6 + 5 = -4x^2 + 24x - 1\)

Set \(gf(x) + 1 = 0\):

\(-4x^2 + 24x - 1 + 1 = 0\)

\(-4x^2 + 24x = 0\)

Factor out \(-4x\):

\(-4x(x - 6) = 0\)

Thus, \(x = 0\) or \(x = 6\).

(i) To express \(x^2 - 4x + 7\) in the form \((x + a)^2 + b\), complete the square:

\(x^2 - 4x + 7 = (x-2)^2 - 4 + 7 = (x-2)^2 + 3\).

(ii) The function \(f(x) = x^2 - 4x + 7\) is decreasing when its derivative \(f'(x) = 2x - 4\) is negative. Solving \(2x - 4 < 0\) gives \(x < 2\). Therefore, the largest value of \(k\) is 2, so \(k \leq 2\).

(iii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\). Solving for \(x\), we get:

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

Since \(x < 1\), choose the negative root: \(x = 2 - \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 - \sqrt{x-3}\) for \(x > 4\).

(iv) The composition \(gf(x)\) is given by:

\(gf(x) = g(f(x)) = \frac{2}{x^2 - 4x + 7 - 1} = \frac{2}{(x-2)^2 + 2}\).

The range of \(gf(x)\) is determined by the values of \((x-2)^2 + 2\), which is always greater than 2. Therefore, the range of \(gf(x)\) is \(0 < gf(x) < \frac{2}{3}\).

(i) Start with \(y = \frac{2}{x^2 - 1}\). Rearrange to get \(x^2 = \frac{2}{y} + 1\).

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

Since \(x < -1\), we take the negative root: \(x = -\sqrt{\frac{2}{y} + 1}\).

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

(ii) Solve \(gf(x) = 5\), which means \(g\left(\frac{2}{x^2 - 1}\right) = 5\).

Substitute \(g(x) = x^2 + 1\): \(\left(\frac{2}{x^2 - 1}\right)^2 + 1 = 5\).

This simplifies to \(\left(\frac{2}{x^2 - 1}\right)^2 = 4\).

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

For \(\frac{2}{x^2 - 1} = 2\), solve \(x^2 - 1 = 1\), giving \(x^2 = 2\) or \(x = \pm \sqrt{2}\).

For \(\frac{2}{x^2 - 1} = -2\), solve \(x^2 - 1 = -1\), giving \(x^2 = 0\) or \(x = 0\).

Since \(x < -1\), the valid solutions are \(x = -\sqrt{2}\) or \(x = -1.41\) only.