(i) To find the equation of the tangent, we first need the derivative of the curve \(y = (3 - 2x)^3\).

Using the chain rule, \(\frac{dy}{dx} = 3(3 - 2x)^2 \times (-2) = -6(3 - 2x)^2\).

At \(x = \frac{1}{2}\), \(\frac{dy}{dx} = -6(3 - 2 \times \frac{1}{2})^2 = -6(2)^2 = -24\).

The slope of the tangent is \(-24\). Using the point-slope form, \(y - 8 = -24(x - \frac{1}{2})\).

Simplifying gives \(y = -24x + 20\).

(ii) To find the area of the shaded region, calculate the area under the curve and subtract the area under the tangent.

Area under the curve from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (3 - 2x)^3 \, dx = \left[ \frac{(3 - 2x)^4}{-8} \right]_0^{1/2} = \left[ \frac{(2)^4}{-8} - \frac{(3)^4}{-8} \right] = \frac{81}{8}\).

Area under the tangent from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (-24x + 20) \, dx = \left[ -12x^2 + 20x \right]_0^{1/2} = -3 + 10 = 7\).

Shaded area = \(\frac{81}{8} - 7 = \frac{9}{8}\) or 1.125.