(a) To find the area between the curves, integrate the difference of the functions from \(x = 0\) to \(x = 4\):

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

Calculate the integrals:

\(\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}}\)

\(\int \left(\frac{1}{2}x^2 - x + 1\right) \, dx = \frac{x^3}{6} - \frac{x^2}{2} + x\)

Evaluate from 0 to 4:

\(\left(\frac{32}{3} + 32 - 0\right) - \left(\frac{44}{3} - 20 + 0\right) = 8\)

(b) Differentiate both curves to find the gradients at \(x = 4\):

Upper curve: \(\frac{dy}{dx} = x^{-\frac{1}{2}}\), at \(x = 4\), gradient = \(\frac{1}{2}\).

Lower curve: \(\frac{dy}{dx} = x - 1\), at \(x = 4\), gradient = 3.

Find \(\alpha\) using the inverse tangent:

\(\alpha = \arctan(3) - \arctan\left(\frac{1}{2}\right)\)

\(\alpha = 45^\circ\)