(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:

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