(a) Differentiate the curve: \(\frac{dy}{dx} = \frac{1}{2} + \frac{1}{3(x-2)^{\frac{4}{3}}}\).

Evaluate \(\frac{dy}{dx}\) at \(x = 3\): \(\frac{1}{2} + \frac{1}{3(3-2)^{\frac{4}{3}}} = \frac{5}{6}\).

The gradient of the normal is the negative reciprocal: \(-\frac{6}{5}\).

Equation of the normal: \(y - \frac{6}{5} = -\frac{6}{5}(x - 3)\).

Simplify: \(y = -\frac{6}{5}x + \frac{24}{5}\).

Set \(y = 0\) to find \(x\): \(0 = -\frac{6}{5}x + \frac{24}{5} \Rightarrow x = 4\).

(b) Integrate the curve from \(x = 2.5\) to \(x = 3\):

\(\int \left( \frac{1}{2}x + \frac{7}{10} - \frac{1}{(x-2)^{\frac{1}{3}}} \right) \, dx\).

Calculate the definite integral: \(\left[ \frac{1}{4}x^2 + \frac{7}{10}x - 3(x-2)^{\frac{2}{3}} \right]_{2.5}^{3}\).

Evaluate: \(\left( \frac{9}{4} + 2.1 - \frac{3}{2} \right) - \left( \frac{6.25}{4} + 1.75 - 3 \times 0.5^{\frac{2}{3}} \right) = 0.48\).

Area of triangle: \(0.6\).

Total area: \(0.48 + 0.6 = 1.08\).