(i) Let \(u = \\cos x\), then \(du = -\\sin x \, dx\). The integral becomes:

\(\int \\sin^3 x \\sqrt{\\cos x} \, dx = \int -u^{1/2} (1-u^2) \, du\).

Integrate to get:

\(-\left( \frac{2}{3} u^{3/2} + \frac{2}{7} u^{7/2} \right)\).

Change limits: when \(x = 0, u = 1\) and when \(x = \frac{1}{2} \pi, u = 0\).

Substitute limits:

\(-\left( \frac{2}{3} (0)^{3/2} + \frac{2}{7} (0)^{7/2} \right) + \left( \frac{2}{3} (1)^{3/2} + \frac{2}{7} (1)^{7/2} \right) = \frac{8}{21}\).

(ii) Differentiate \(y = \\sin^3 x \\sqrt{\\cos x}\) using the product and chain rules:

\(\frac{dy}{dx} = 3 \\sin^2 x \\cos x \\sqrt{\\cos x} - \frac{1}{2} \\sin^3 x \\frac{\\sin x}{\\sqrt{\\cos x}}\).

Set \(\frac{dy}{dx} = 0\) and simplify:

\(3 \\sin^2 x \\cos^2 x = \frac{1}{2} \\sin^4 x\).

Rearrange to find \(\tan^2 x = 6\).

Thus, \(x = \arctan(\\sqrt{6}) \approx 1.183\).