(i) To find the \(x\)-coordinate of \(Q\), we first find the derivative of the curve \(y = \frac{3}{\sqrt{1 + 4x}}\). The derivative is \(\frac{dy}{dx} = 3 \times -\frac{1}{2} \times (1 + 4x)^{-\frac{3}{2}} \times 4\).

At \(x = 2\), the gradient \(m\) is \(-\frac{2}{9}\). The perpendicular gradient (normal) is \(\frac{9}{2}\).

The equation of the normal is \(y - 1 = \frac{9}{2}(x - 2)\).

Setting \(y = 0\) to find \(Q\), we solve \(0 - 1 = \frac{9}{2}(x - 2)\), giving \(x = \frac{16}{9}\).

(ii) The area under the curve from \(x = 0\) to \(x = 2\) is \(\int_{0}^{2} \frac{3}{\sqrt{1 + 4x}} \, dx = \frac{3}{2} \sqrt{1 + 4x} \bigg|_0^2 \times \frac{1}{4}\).

Evaluating gives \(3\).

The area of the triangle is \(\frac{1}{2} \times 1 \times \frac{2}{9} = \frac{1}{9}\).

The shaded area is \(3 - \frac{1}{9} = \frac{8}{9}\).