(a) To find the value of \(a\), we need to find the maximum point of the curve \(y = \\sin x \\cos 2x\). Differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \\cos x \\cos 2x - 2 \\sin x \\sin 2x\).

Using the double angle formula, \(\\sin 2x = 2 \\sin x \\cos x\), we can rewrite the derivative as:

\(\frac{dy}{dx} = \\cos x (1 - 6 \\sin^2 x) = 0\).

Setting \(\frac{dy}{dx} = 0\), we solve for \(x\):

\(\\cos x (1 - 6 \\sin^2 x) = 0\).

This gives \(x = 0.42\) as the maximum point, so \(a = 0.42\).

(b) To find the exact area of the region \(R\), we integrate \(y = \\sin x \\cos 2x\) from \(\frac{1}{4} \\pi\) to \(\frac{3}{4} \\pi\):

Using the double angle formula, \(y = \frac{1}{2} (\\sin 3x + \\sin x)\).

The integral becomes:

\(\int_{\frac{1}{4} \\pi}^{\frac{3}{4} \\pi} \frac{1}{2} (\\sin 3x + \\sin x) \, dx\).

Calculate the integral:

\(\frac{1}{2} \left[ -\frac{1}{3} \\cos 3x - \\cos x \right]_{\frac{1}{4} \\pi}^{\frac{3}{4} \\pi}\).

Evaluating the limits gives:

\(\frac{2\sqrt{2}}{3}\).