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

\(0 = 3 + \frac{12}{2-x}\)

\(-3 = \frac{12}{2-x}\)

\(-3(2-x) = 12\)

\(-6 + 3x = 12\)

\(3x = 18\)

\(x = 6\)

The curve crosses the x-axis at \((6, 0)\).

Find \(\frac{dy}{dx}\):

\(y = 3 + \frac{12}{2-x}\)

\(\frac{dy}{dx} = 0 + 12(2-x)^{-2}(-1)\)

\(\frac{dy}{dx} = \frac{12}{(2-x)^2}\)

At \(x = 6\), \(\frac{dy}{dx} = \frac{12}{(2-6)^2} = \frac{12}{16} = \frac{3}{4}\)

The equation of the tangent is \(y = \frac{1}{4}(x - 6)\) or \(4y = 3x - 18\).

(ii) Given \(\frac{dx}{dt} = 0.04\) and \(x = 4\), find \(\frac{dy}{dt}\):

\(\frac{dy}{dx} = 3\) when \(x = 4\).

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 3 \times 0.04 = 0.12\)