First, calculate fg(x):

fg(x) = f(g(x)) = -3(-x² - 1)² + 2 = -3(x⁴ + 2x² + 1) + 2 = -3x⁴ - 6x² - 3 + 2 = -3x⁴ - 6x² - 1.

Next, calculate gf(x):

gf(x) = g(f(x)) = -(-3x² + 2)² - 1 = -(9x⁴ - 12x² + 4) - 1 = -9x⁴ + 12x² - 5.

Substitute into the equation:

fg(x) - gf(x) + 8 = (-3x⁴ - 6x² - 1) - (-9x⁴ + 12x² - 5) + 8 = 6x⁴ - 18x² + 12 = 0.

Factor the equation:

\(6(x⁴ - 3x² + 2) = 0.\)

\(Let y = x², then y² - 3y + 2 = 0.\)

\(Factor the quadratic: (y - 1)(y - 2) = 0.\)

\(So, y = 1 or y = 2.\)

\(Thus, x² = 1 or x² = 2.\)

\(Therefore, x = -1 or x = -\sqrt{2}.\)

(a) To find \(gf(x)\), substitute \(g(x) = ax + 1\) into \(f(x)\):

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

(b) Given \(gf(2) = 11\), substitute \(x = 2\) into \(gf(x)\):

\(gf(2) = a\left(2 + \frac{1}{2}\right) + 1 = 11\).

\(a\left(2.5\right) + 1 = 11\)

\(2.5a + 1 = 11\)

\(2.5a = 10\)

\(a = 4\).

(c) The function \(f(x) = x + \frac{1}{x}\) is not one-one because it has a minimum point at \(x = 1\), meaning it is not strictly increasing or decreasing. Therefore, \(f\) does not have an inverse.

(d) To find \(g^{-1}(x)\), solve \(y = ax + 1\) for \(x\):

\(y - 1 = ax\)

\(x = \frac{y - 1}{a}\)

For \(a = 5\), \(g^{-1}(x) = \frac{x - 1}{5}\).

Substitute \(f(x) = x + \frac{1}{x}\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{x + \frac{1}{x} - 1}{5}\).

(e) The composite function \(fg\) cannot be formed because the domain of \(f\) (\(x > 0\)) does not include the whole range of \(g\), which is all real numbers.

To find \(fg(x)\), we substitute \(g(x) = 2x + 4\) into \(f(x)\).

\(fg(x) = f(2x + 4) = 2((2x + 4)^2) - 16(2x + 4) + 23\).

First, calculate \((2x + 4)^2 = 4x^2 + 16x + 16\).

Then, \(2(4x^2 + 16x + 16) = 8x^2 + 32x + 32\).

Next, calculate \(-16(2x + 4) = -32x - 64\).

Combine these results: \(8x^2 + 32x + 32 - 32x - 64 + 23\).

Simplify: \(8x^2 + 32 - 64 + 23 = 8x^2 - 9\).

(a) The domain of \(f^{-1}\) is determined by the range of \(f(x)\). Since \(f(x)\) is undefined at \(x = \frac{1}{2}\), the domain of \(f^{-1}\) is \(x \neq 1\) or \(x < 1\), \(x > 1\) or \((-\infty, 1) \cup (1, \infty)\).

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

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

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

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

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

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

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

(c) Substitute \(x = 3\) into \(f^{-1}(x)\):

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

Then, \(g(f^{-1}(3)) = g(2) = 2^2 + 4 = 8\).

(d) \(g(x) = x^2 + 4\) is not one-to-one because it is a quadratic function, which means \(g^{-1}(x)\) cannot be found.

(e) Show that \(1 + \frac{2}{2x-1} = \frac{2x+1}{2x-1}\):

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

Find the equation of the tangent at \(x = 1\):

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

The derivative \(f'(x) = \frac{-4}{(2x-1)^2}\) gives \(f'(1) = -4\)

The equation of the tangent is \(y - 3 = -4(x - 1)\) or \(y = -4x + 7\)

The tangent crosses the x-axis at \(x = \frac{7}{4}\) and the y-axis at \(y = 7\).

The area of the triangle is \(\frac{1}{2} \times \frac{7}{4} \times 7 = \frac{49}{8}\).