(i) Let \(u = \\sin 2x\), then \(du = 2 \\cos 2x \, dx\) or \(dx = \frac{du}{2 \\cos 2x}\).

The integrand becomes \(\\sin^3 2x \\cos^3 2x = u^3 (1-u^2)^{3/2}\).

Substitute and simplify: \(\\int u^3 (1-u^2)^{3/2} \, du\).

Change limits: when \(x = 0, u = 0\) and when \(x = \frac{\\pi}{4}, u = 1\).

Integrate: \(\\int_0^1 \frac{1}{2} u^3 (1-u^2) \, du = \frac{1}{2} \\int_0^1 (u^3 - u^5) \, du\).

Calculate: \(\frac{1}{2} \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_0^1 = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right) = \frac{1}{24}\).

Thus, \(A = \frac{1}{24}\).

(ii) Given \(\\int_0^{k\\pi} |\\sin^3 2x \\cos^3 2x| \, dx = 40A\), substitute \(A = \frac{1}{24}\).

\(40 \times \frac{1}{24} = \frac{5}{3}\).

Since the integral over one period \([0, \frac{\\pi}{2}]\) is \(\frac{1}{24}\), and \(k\) periods are \(\frac{5}{3}\), solve for \(k\):

\(k \times \frac{1}{24} = \frac{5}{3} \Rightarrow k = 10\).