(i) To find the equation of the tangent at B, first find the derivative of \(y = \frac{8}{3x + 2}\). The derivative is \(\frac{dy}{dx} = -\frac{24}{(3x + 2)^2}\). At B (2, 1), \(\frac{dy}{dx} = -\frac{3}{8}\).

The equation of the tangent line at B is \(y - 1 = -\frac{3}{8}(x - 2)\).

To find where this line crosses the x-axis, set \(y = 0\):

\(0 - 1 = -\frac{3}{8}(x - 2)\)

\(-1 = -\frac{3}{8}x + \frac{3}{4}\)

\(-\frac{7}{4} = -\frac{3}{8}x\)

\(x = \frac{7}{4} \times \frac{8}{3} = \frac{14}{3}\)

The area of triangle BDC is \(\frac{1}{2} \times 2 \times \frac{2}{3} \times 1 = \frac{4}{3}\).

(ii) The volume of the solid formed by rotating the region ODBA about the x-axis is given by:

\(V = \pi \int_0^2 y^2 \, dx = \pi \int_0^2 \left(\frac{8}{3x + 2}\right)^2 \, dx\)

\(= \pi \int_0^2 \frac{64}{(3x + 2)^2} \, dx\)

Let \(u = 3x + 2\), then \(du = 3 \, dx\), so \(dx = \frac{1}{3} \, du\).

Change the limits: when \(x = 0, u = 2\); when \(x = 2, u = 8\).

\(V = \pi \int_2^8 \frac{64}{u^2} \times \frac{1}{3} \, du\)

\(= \frac{64\pi}{3} \int_2^8 u^{-2} \, du\)

\(= \frac{64\pi}{3} \left[ -u^{-1} \right]_2^8\)

\(= \frac{64\pi}{3} \left( -\frac{1}{8} + \frac{1}{2} \right)\)

\(= \frac{64\pi}{3} \times \frac{3}{8}\)

\(= 8\pi\).