(i) To find the volume of the solid obtained by rotating the region about the x-axis, we use the formula:

\(V = \pi \int_0^3 (x^3 + x^2) \, dx\)

Integrating, we have:

\(\pi \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^3\)

\(= \pi \left[ \frac{81}{4} + 9 - 0 \right]\)

\(= \frac{117\pi}{4}\)

(ii) First, find the derivative \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{2} (x^3 + x^2)^{-1/2} \times (3x^2 + 2x)\)

At \(x = 3\), \(y = 6\).

The gradient \(m\) at \(x = 3\) is:

\(m = \frac{1}{2} \times \frac{1}{6} \times 33 = \frac{11}{4}\)

The equation of the normal is:

\(y - 6 = -\frac{4}{11}(x - 3)\)

When \(x = 0\), \(y =7 \frac{1}{11}\).