(i) To evaluate \(fg(2)\), first find \(g(2)\):

\(g(2) = 2(2 - 1)^3 + 8 = 2(1)^3 + 8 = 2 + 8 = 10\).

Then, find \(f(g(2)) = f(10)\):

\(f(10) = 3(10) - 4 = 30 - 4 = 26\).

Therefore, \(fg(2) = 26\).

(iv) To find \(f^{-1}(x)\), solve \(y = 3x - 4\) for \(x\):

\(y + 4 = 3x\)

\(x = \frac{y + 4}{3}\)

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

To find \(g^{-1}(x)\), solve \(y = 2(x - 1)^3 + 8\) for \(x\):

\(y - 8 = 2(x - 1)^3\)

\(\frac{y - 8}{2} = (x - 1)^3\)

\(x - 1 = \sqrt[3]{\frac{y - 8}{2}}\)

\(x = \sqrt[3]{\frac{y - 8}{2}} + 1\)

Thus, \(g^{-1}(x) = \sqrt[3]{\frac{x - 8}{2}} + 1\).

(i) To find \(ff(x)\), we substitute \(f(x)\) into itself:

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

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

Simplifying the numerator:

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

Simplifying the denominator:

\(2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - 1 = \frac{2x+6 - (2x-1)}{2x-1} = \frac{7}{2x-1}\)

Thus, \(ff(x) = \frac{7x}{7} = x\).

(ii) To find \(f^{-1}(x)\), we set \(y = \frac{x+3}{2x-1}\) and solve for \(x\):

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

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

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

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

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

(i) To find \(fg(x)\), substitute \(g(x) = x^2 - 2\) into \(f(x) = 2x + 1\):

\(fg(x) = f(g(x)) = f(x^2 - 2) = 2(x^2 - 2) + 1 = 2x^2 - 4 + 1 = 2x^2 - 3\).

To find \(gf(x)\), substitute \(f(x) = 2x + 1\) into \(g(x) = x^2 - 2\):

\(gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 - 2 = 4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1\).

(ii) Set \(fg(a) = gf(a)\):

\(2a^2 - 3 = 4a^2 + 4a - 1\)

\(2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0\)

\((a + 1)^2 = 0 \Rightarrow a = -1\).

(iii) Solve \(g(b) = b\):

\(b^2 - 2 = b \Rightarrow b^2 - b - 2 = 0\)

\((b + 1)(b - 2) = 0 \Rightarrow b = 2\) (also allow \(b = -1\)).

(iv) Find \(f^{-1}(x)\):

\(y = 2x + 1 \Rightarrow x = \frac{1}{2}(y - 1) \Rightarrow f^{-1}(x) = \frac{1}{2}(x - 1)\).

Substitute into \(g(x)\):

\(f^{-1}g(x) = \frac{1}{2}(x^2 - 3)\).

(v) Solve \(h(x) = x^2 - 2\) for \(x \leq 0\):

\(y = x^2 - 2 \Rightarrow x^2 = y + 2 \Rightarrow x = -\sqrt{y + 2} \Rightarrow h^{-1}(x) = -\sqrt{x + 2}\).

First, substitute \(g(x) = x + 3\) into \(f(x)\):

\(fg(x) = 2 - \frac{5}{(x+3)+2} = 2 - \frac{5}{x+5}\).

To combine into a single fraction, express \(2\) as \(\frac{2(x+5)}{x+5}\):

\(fg(x) = \frac{2(x+5)}{x+5} - \frac{5}{x+5}\).

Combine the fractions:

\(fg(x) = \frac{2(x+5) - 5}{x+5}\).

Simplify the numerator:

\(2(x+5) - 5 = 2x + 10 - 5 = 2x + 5\).

Thus, \(fg(x) = \frac{2x+5}{x+5}\).

First, substitute \(f(x) = 2x + 3\) into \(g(x) = x^2 - 2x\):

\(gf(x) = (2x + 3)^2 - 2(2x + 3)\).

Expand \((2x + 3)^2\):

\((2x + 3)^2 = 4x^2 + 12x + 9\).

Expand \(-2(2x + 3)\):

\(-2(2x + 3) = -4x - 6\).

Combine the expressions:

\(gf(x) = 4x^2 + 12x + 9 - 4x - 6\).

Simplify:

\(gf(x) = 4x^2 + 8x + 3\).

Complete the square for \(4x^2 + 8x + 3\):

Factor out 4 from the quadratic terms:

\(4(x^2 + 2x) + 3\).

Complete the square inside the parentheses:

\(x^2 + 2x = (x + 1)^2 - 1\).

Substitute back:

\(4((x + 1)^2 - 1) + 3\).

Expand and simplify:

\(4(x + 1)^2 - 4 + 3 = 4(x + 1)^2 - 1\).

Thus, \(gf(x) = 4(x + 1)^2 - 1\).