(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:

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

(a) To find the minimum point \(M\), we first find the derivative of \(y = x^3 \ln x\) using the product rule:

\(\frac{d}{dx}(x^3 \ln x) = x^3 \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(x^3)\).

This simplifies to \(\frac{x^3}{x} + 3x^2 \ln x = x^2 + 3x^2 \ln x\).

Setting the derivative to zero gives:

\(x^2(1 + 3 \ln x) = 0\).

Since \(x > 0\), we solve \(1 + 3 \ln x = 0\) to find \(x = e^{-\frac{1}{3}} = \frac{1}{\sqrt{e}}\).

Substituting back into the original equation, \(y = \left( \frac{1}{\sqrt{e}} \right)^3 \ln \left( \frac{1}{\sqrt{e}} \right) = -\frac{1}{3e}\).

Thus, the coordinates of \(M\) are \(\left( \frac{1}{\sqrt{e}}, -\frac{1}{3e} \right)\).

(b) To find the area of the shaded region, we integrate \(y = x^3 \ln x\) from \(x = \frac{1}{2}\) to \(x = 1\).

Using integration by parts, let \(u = \ln x\) and \(dv = x^3 dx\), then \(du = \frac{1}{x} dx\) and \(v = \frac{x^4}{4}\).

The integral becomes:

\(\int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx\).

This simplifies to:

\(\frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx = \frac{x^4}{4} \ln x - \frac{x^4}{16}\).

Evaluating from \(x = \frac{1}{2}\) to \(x = 1\):

\(\left[ \frac{1^4}{4} \ln 1 - \frac{1^4}{16} \right] - \left[ \frac{(\frac{1}{2})^4}{4} \ln \frac{1}{2} - \frac{(\frac{1}{2})^4}{16} \right]\).

This simplifies to:

\(0 - \frac{1}{16} - \left( \frac{1}{64} \ln \frac{1}{2} - \frac{1}{256} \right)\).

Further simplification gives:

\(\frac{15}{256} - \frac{1}{64} \ln 2\).

(a) To find the coordinates of \(M\), we need to find the minimum point of the curve \(y = (3-x)e^{-\frac{1}{3}x}\).

First, find the derivative \(\frac{dy}{dx}\) using the product rule:

\(\frac{dy}{dx} = -e^{-\frac{1}{3}x} - \frac{1}{3}(3-x)e^{-\frac{1}{3}x}\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(-e^{-\frac{1}{3}x} - \frac{1}{3}(3-x)e^{-\frac{1}{3}x} = 0\).

Simplify and solve for \(x\):

\(x = 6\).

Substitute \(x = 6\) back into the original equation to find \(y\):

\(y = (3-6)e^{-\frac{1}{3} \times 6} = -3e^{-2}\).

Thus, the coordinates of \(M\) are \((6, -3e^{-2})\).

(b) To find the area of the shaded region, integrate the function from \(x = 0\) to \(x = 3\):

\(\int_{0}^{3} (3-x)e^{-\frac{1}{3}x} \, dx\).

Using integration by parts, we find:

\(-3e^{-\frac{1}{3}x}(3-x) + 9e^{-\frac{1}{3}x}\).

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[-3e^{-1}(3-3) + 9e^{-1}\right] - \left[-3e^{0}(3-0) + 9e^{0}\right] = \frac{9}{e}\).

Thus, the area of the shaded region is \(\frac{9}{e}\).

(a) To find the coordinates of \(M\), we first find the derivative of \(y = (2-x)e^{-\frac{1}{2}x}\) using the product rule:

\(\frac{dy}{dx} = \left(-\frac{1}{2}(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x}\right)\).

Set \(\frac{dy}{dx} = 0\) to find the critical points:

\(-\frac{1}{2}(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x} = 0\).

Simplifying gives \(x = 4\).

Substitute \(x = 4\) back into the original equation to find \(y\):

\(y = (2-4)e^{-\frac{1}{2} \times 4} = -2e^{-2}\).

Thus, the coordinates of \(M\) are \((4, -2e^{-2})\).

(b) To find the area of the shaded region, integrate the function from \(x = 0\) to \(x = 2\):

\(\int_{0}^{2} (2-x)e^{-\frac{1}{2}x} \, dx\).

Using integration by parts, we find:

\(\int (2-x)e^{-\frac{1}{2}x} \, dx = -2(2-x)e^{-\frac{1}{2}x} - 2xe^{-\frac{1}{2}x}\).

Evaluate from \(x = 0\) to \(x = 2\):

\(\left[-2(2-x)e^{-\frac{1}{2}x} - 2xe^{-\frac{1}{2}x}\right]_{0}^{2} = 4e^{-1}\).

Thus, the area of the shaded region is \(4e^{-1}\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the derivative of \(y = (x + 1) e^{-\frac{1}{3}x}\) and set it to zero to find the critical points.

Using the product rule, the derivative is:

\(\frac{dy}{dx} = -\frac{1}{3}(x + 1)e^{-\frac{1}{3}x} + e^{-\frac{1}{3}x}\)

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

\(-\frac{1}{3}(x + 1) + 1 = 0\)

\(-\frac{1}{3}x - \frac{1}{3} + 1 = 0\)

\(-\frac{1}{3}x + \frac{2}{3} = 0\)

\(x = 2\)

(ii) To find the area of the shaded region, we integrate \(y = (x + 1) e^{-\frac{1}{3}x}\) from \(x = -1\) to \(x = 0\).

Using integration by parts, we have:

\(\int (x + 1) e^{-\frac{1}{3}x} \, dx = -3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\)

Evaluating from \(x = -1\) to \(x = 0\):

\(\left[-3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\right]_{-1}^{0}\)

\(= \left[-3(0 + 1)e^{0} - 9e^{0}\right] - \left[-3(-1 + 1)e^{\frac{1}{3}} - 9e^{\frac{1}{3}}\right]\)

\(= [-3 - 9] - [0 - 9e^{-\frac{1}{3}}]\)

\(= -12 + 9e^{-\frac{1}{3}}\)

\(= 9e^{-3} - 12\)