(i) The curve cuts the y-axis at \(A = (0, 1)\) and the line \(x = 5\) at \(B = (5, \frac{1}{2})\).

The slope of line \(AB\) is \(\frac{\frac{1}{2} - 1}{5 - 0} = -\frac{1}{10}\).

Using the point-slope form: \(y - 1 = -\frac{1}{10}(x - 0)\).

Simplifying gives: \(y = -\frac{1}{10}x + 1\).

(ii) To find the volume, calculate the volume of revolution for both the curve and the line, then subtract.

For the curve: \(V_{\text{curve}} = \pi \int_{0}^{5} \left( \frac{1}{(3x+1)^{\frac{1}{4}}} \right)^2 \, dx\).

\(= \pi \int_{0}^{5} (3x+1)^{-\frac{1}{2}} \, dx\).

Integrating gives: \(\frac{2\pi}{3} [(3x+1)^{\frac{1}{2}}]_{0}^{5}\).

\(= \frac{2\pi}{3} [4 - 1] = 2\pi\).

For the line: \(V_{\text{line}} = \pi \int_{0}^{5} \left( -\frac{1}{10}x + 1 \right)^2 \, dx\).

\(= \pi \int_{0}^{5} \left( \frac{1}{100}x^2 - \frac{1}{5}x + 1 \right) \, dx\).

Integrating gives: \(\pi \left[ \frac{1}{300}x^3 - \frac{1}{10}x^2 + x \right]_{0}^{5}\).

\(= \pi \left[ \frac{125}{300} - \frac{25}{10} + 5 \right] = \frac{35\pi}{12}\).

The volume of the shaded region is \(V_{\text{line}} - V_{\text{curve}} = \frac{35\pi}{12} - 2\pi = \frac{11\pi}{12}\).

(a) To find the coordinates of \(A\), substitute \(x = 0\) into the circle equation \((x-2)^2 + y^2 = 8\):

\((-2)^2 + y^2 = 8\)

\(4 + y^2 = 8\)

\(y^2 = 4\)

\(y = 2\) (since \(A\) is on the positive \(y\)-axis)

Thus, \(A = (0, 2)\).

For \(B\), since \(AB\) is parallel to the \(x\)-axis, \(y = 2\). Substitute \(y = 2\) into the circle equation:

\((x-2)^2 + 4 = 8\)

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

\(x-2 = \pm 2\)

\(x = 4\) (since \(B\) is on the right side of the circle)

Thus, \(B = (4, 2)\).

(b) The volume of revolution is found by integrating the area of the segment rotated about the \(x\)-axis:

Volume = \(\pi \int_0^4 (8 - (x-2)^2) \, dx\)

\(= \pi \left[ 8x - \frac{(x-2)^3}{3} \right]_0^4\)

\(= \pi \left[ 32 - \frac{16}{3} \right]\)

Volume of cylinder = \(\pi \times 2^2 \times 4 = 16\pi\)

Total volume = \(26\frac{2}{3}\pi - 16\pi = 10\frac{2}{3}\pi\)

The volume \(V\) of the solid of revolution is given by the formula:

\(V = \pi \int_0^1 y^2 \, dx\)

Substitute \(y = \frac{9}{2-x}\) into the formula:

\(V = \pi \int_0^1 \left(\frac{9}{2-x}\right)^2 \, dx\)

\(V = \pi \int_0^1 \frac{81}{(2-x)^2} \, dx\)

Integrate \(\frac{81}{(2-x)^2}\):

\(\int \frac{81}{(2-x)^2} \, dx = -81 \left(2-x\right)^{-1} \times (-1)\)

\(= 81 \left(2-x\right)^{-1}\)

Evaluate from 0 to 1:

\(V = \pi \left[ 81 \left(2-1\right)^{-1} - 81 \left(2-0\right)^{-1} \right]\)

\(= \pi \left[ 81 \times 1 - 81 \times \frac{1}{2} \right]\)

\(= \pi \left[ 81 - 40.5 \right]\)

\(= \pi \times 40.5\)

\(= \frac{81\pi}{2}\)

(i) To find the coordinates of \(A\) and \(B\), set \(y = x + \frac{4}{x} = 5\). Solving for \(x\), we get:

\(x + \frac{4}{x} = 5\)

\(x^2 - 5x + 4 = 0\)

\((x-1)(x-4) = 0\)

Thus, \(x = 1\) or \(x = 4\). Therefore, \(A(1, 5)\) and \(B(4, 5)\).

To find \(M\), differentiate \(y = x + \frac{4}{x}\):

\(\frac{dy}{dx} = 1 - \frac{4}{x^2}\)

Set \(\frac{dy}{dx} = 0\) to find the minimum point:

\(1 - \frac{4}{x^2} = 0\)

\(x^2 = 4\)

\(x = 2\)

Substitute \(x = 2\) into \(y = x + \frac{4}{x}\):

\(y = 2 + \frac{4}{2} = 4\)

Thus, \(M(2, 4)\).

(ii) The volume of the solid of revolution is given by:

\(V = \pi \int_{1}^{4} y^2 \, dx\)

\(y = x + \frac{4}{x}\), so \(y^2 = (x + \frac{4}{x})^2 = x^2 + 8 + \frac{16}{x^2}\)

Integrate:

\(\int (x^2 + 8 + \frac{16}{x^2}) \, dx = \frac{x^3}{3} + 8x - \frac{16}{x}\)

Evaluate from 1 to 4:

\(\left[ \frac{x^3}{3} + 8x - \frac{16}{x} \right]_{1}^{4} = \left( \frac{64}{3} + 32 - 4 \right) - \left( \frac{1}{3} + 8 - 16 \right)\)

\(= \left( \frac{64}{3} + 28 \right) - \left( \frac{1}{3} - 8 \right)\)

\(= \frac{64}{3} + 28 + 8 - \frac{1}{3}\)

\(= \frac{63}{3} + 36\)

\(= 21 + 36 = 57\)

Volume of cylinder: \(\pi \times 5^2 \times 3 = 75\pi\)

Volume of solid: \(75\pi - 57\pi = 18\pi\)

The volume of the solid of revolution formed by rotating the curve \(y = \frac{a}{x}\) from \(x = 1\) to \(x = 3\) about the x-axis is given by:

\(V = \pi \int_{1}^{3} \left( \frac{a}{x} \right)^2 \, dx\)

\(= \pi \int_{1}^{3} \frac{a^2}{x^2} \, dx\)

Integrating \(\frac{a^2}{x^2}\) gives:

\(= \pi \left[ -\frac{a^2}{x} \right]_{1}^{3}\)

\(= \pi \left( -\frac{a^2}{3} + \frac{a^2}{1} \right)\)

\(= \pi \left( a^2 - \frac{a^2}{3} \right)\)

\(= \pi \left( \frac{2a^2}{3} \right)\)

We are given that this volume is \(24\pi\), so:

\(\pi \left( \frac{2a^2}{3} \right) = 24\pi\)

\(\frac{2a^2}{3} = 24\)

\(2a^2 = 72\)

\(a^2 = 36\)

\(a = 6\)