(i) To find the equation of the normal, first find the gradient of the tangent at \(P(3, 3)\). The gradient \(m\) is given by \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). At \(x = 3\), \(\frac{dy}{dx} = \frac{6}{\sqrt{4(3) - 3}} = 2\). The gradient of the normal is the negative reciprocal, \(m = -\frac{1}{2}\). The equation of the normal is \(y - 3 = -\frac{1}{2}(x - 3)\), which simplifies to \(x + 2y = 9\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). Let \(u = 4x - 3\), then \(du = 4dx\), so \(dx = \frac{1}{4}du\). The integral becomes \(\int \frac{6}{\sqrt{u}} \cdot \frac{1}{4} \, du = \frac{3}{2} \int u^{-1/2} \, du\). This integrates to \(3u^{1/2} + c\). Substituting back \(u = 4x - 3\), we get \(y = 3(4x - 3)^{1/2} + c\). Using the point \((3, 3)\), substitute to find \(c\): \(3 = 3(4(3) - 3)^{1/2} + c\), which gives \(c = -6\). Therefore, the equation of the curve is \(y = 3(4x - 3)^{1/2} - 6\).