(a) To find the area between the curves, we calculate the definite integral of the difference between the two functions from \(x = \frac{1}{4}\) to \(x = 4\):

\(\int_{1/4}^{4} \left( \frac{5}{2} - x^{-\frac{1}{2}} - x^{-\frac{1}{2}} \right) \, dx\)

\(= \int_{1/4}^{4} \left( \frac{5}{2} - 2x^{-\frac{1}{2}} \right) \, dx\)

Integrating term by term, we have:

\(\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}} \bigg|_{1/4}^{4}\)

Substituting the limits:

\(\left( \frac{5}{2} \times 4 - \frac{2}{3} \times 4^{\frac{3}{2}} \right) - \left( \frac{5}{2} \times \frac{1}{4} - \frac{2}{3} \times \left( \frac{1}{4} \right)^{\frac{3}{2}} \right)\)

\(= \left( 10 - \frac{16}{3} \right) - \left( \frac{5}{8} - \frac{1}{12} \right)\)

\(= \frac{9}{8}\)

(b) The derivative of \(y = x^{-\frac{1}{2}}\) is \(\frac{dy}{dx} = -\frac{1}{2}x^{-\frac{3}{2}}\).

At \(x = 1\), the gradient \(m = -\frac{1}{2}\).

The equation of the normal at \((1, 1)\) is:

\(y - 1 = 2(x - 1)\)

Substituting \(x = 0\) to find \(p\):

\(y - 1 = 2(0 - 1)\)

\(y - 1 = -2\)

\(y = -1\)

Thus, \(p = -1\).