(i) Given \(\frac{d^2y}{dx^2} = -4x\), integrate to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int -4x \, dx = -2x^2 + c\)

Since the curve has a maximum at \(x = 2\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(0 = -2(2)^2 + c \Rightarrow c = 8\)

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (-2x^2 + 8) \, dx = -\frac{2x^3}{3} + 8x + C\)

Use the point (2, 12) to find \(C\):

\(12 = -\frac{2(2)^3}{3} + 8(2) + C\)

\(12 = -\frac{16}{3} + 16 + C\)

\(C = \frac{4}{3}\)

Thus, the equation of the curve is \(y = -\frac{2x^3}{3} + 8x + \frac{4}{3}\).

(ii) To find \(\frac{dy}{dt}\), use the chain rule:

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

From part (i), \(\frac{dy}{dx} = -2x^2 + 8\).

At \(x = 3\), \(\frac{dy}{dx} = -2(3)^2 + 8 = -18 + 8 = -10\).

Given \(\frac{dx}{dt} = 0.05\),

\(\frac{dy}{dt} = -10 \times 0.05 = -0.5\).

Thus, the \(y\)-coordinate is decreasing at 0.5 units per second.