(a) To find the equation of the normal, first find the gradient of the tangent at \(x = 2\) using \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\).

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{2}(2) + \frac{72}{2^4} = 1 + \frac{72}{16} = 1 + 4.5 = 5.5\).

The gradient of the normal is the negative reciprocal: \(-\frac{1}{5.5} = -\frac{2}{11}\).

Using point-slope form \(y - y_1 = m(x - x_1)\) with \(P(2, 8)\):

\(y - 8 = -\frac{2}{11}(x - 2)\).

Simplifying gives \(y = -\frac{2}{11}x + \frac{92}{11}\).

(b) Integrate \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\) to find \(y\):

\(y = \int \left( \frac{1}{2}x + \frac{72}{x^4} \right) dx = \frac{1}{4}x^2 - \frac{24}{x^3} + C\).

Use the point \(P(2, 8)\) to find \(C\):

\(8 = \frac{1}{4}(2)^2 - \frac{24}{2^3} + C\).

\(8 = 1 - 3 + C\).

\(C = 10\).

Thus, the equation of the curve is \(y = \frac{1}{4}x^2 - \frac{24}{x^3} + 10\).