(a) To find the volume generated when the shaded region is rotated about the \(y\)-axis, we use the formula for the volume of revolution:

\(\text{Volume} = \pi \int x^2 \, dy\)

Given \(y = \frac{6}{x}\), we have \(x = \frac{6}{y}\). Therefore, \(x^2 = \left(\frac{6}{y}\right)^2 = \frac{36}{y^2}\).

Substitute into the integral:

\(\text{Volume} = \pi \int \frac{36}{y^2} \, dy\)

\(= \pi \left[ -\frac{36}{y} \right]\)

Evaluate from \(y = 2\) to \(y = 6\):

\(= \pi \left( -\frac{36}{6} + \frac{36}{2} \right)\)

\(= \pi ( -6 + 18 )\)

\(= 12\pi\)

Subtract the volume of the cylinder \(\pi \times 1^2 \times 4 = 4\pi\):

\(\text{Total Volume} = 12\pi - 4\pi = 8\pi\)

(b) The gradient of the curve \(y = \frac{6}{x}\) is given by differentiating:

\(\frac{dy}{dx} = -\frac{6}{x^2}\)

The gradient of the line \(y + 2x = 0\) is \(-2\). Set \(-\frac{6}{x^2} = -2\):

\(\frac{6}{x^2} = 2\)

\(x^2 = 3\)

\(x = \sqrt{3}\)

Substitute \(x = \sqrt{3}\) into \(y = \frac{6}{x}\):

\(y = \frac{6}{\sqrt{3}} = 2\sqrt{3}\)

Check if \(y = 2x\):

\(2\sqrt{3} = 2 \times \sqrt{3}\)

Thus, \(X\) lies on \(y = 2x\).

(a) To find the coordinates of \(A\) and \(B\), solve the simultaneous equations:

\(\frac{8}{x+2} = 4 - \frac{1}{2}x\)

Solving gives \(x = 0\) or \(x = 6\).

For \(x = 0\), \(y = \frac{8}{0+2} = 4\), so \(A(0, 4)\).

For \(x = 6\), \(y = \frac{8}{6+2} = 1\), so \(B(6, 1)\).

At \(C\), the derivative of \(y = \frac{8}{x+2}\) is \(-\frac{8}{(x+2)^2}\).

Set \(-\frac{8}{(x+2)^2} = -\frac{1}{2}\) (slope of \(AB\)) and solve for \(x\):

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

\((x+2)^2 = 16\)

\(x = 2\)

Substitute \(x = 2\) into \(y = \frac{8}{x+2}\) to find \(y = 2\), so \(C(2, 2)\).

(b) The volume under the line is:

\(\pi \int_{0}^{6} \left(-\frac{1}{2}x + 4\right)^2 \, dx = \pi \left[ \frac{x^3}{12} - 2x^2 + 16x \right]_{0}^{6} = 42\pi\)

The volume under the curve is:

\(\pi \int_{0}^{6} \left(\frac{8}{x+2}\right)^2 \, dx = \pi \left[ \frac{-64}{x+2} \right]_{0}^{6} = 24\pi\)

The volume of the shaded region is:

\(42\pi - 24\pi = 18\pi\)

To find the volume of the solid obtained by rotating the region about the \(y\)-axis, we use the method of cylindrical shells. The formula for the volume is:

\(V = \pi \int_{1}^{5} (y - 1) \, dy\)

First, integrate \(y - 1\):

\(\int (y - 1) \, dy = \frac{y^2}{2} - y\)

Apply the limits from 1 to 5:

\(V = \pi \left[ \left( \frac{25}{2} - 5 \right) - \left( \frac{1}{2} - 1 \right) \right]\)

Calculate the expression:

\(V = \pi \left[ \frac{25}{2} - 5 - \frac{1}{2} + 1 \right]\)

\(V = \pi \left[ \frac{25}{2} - \frac{1}{2} - 4 \right]\)

\(V = \pi \left[ \frac{24}{2} - 4 \right]\)

\(V = \pi \times 8\)

\(V = 8\pi\)

Thus, the volume obtained is \(8\pi\) or approximately 25.1.