(i) Use the angle addition formulas: \(\sin(3x + x) = \sin 4x = \sin 3x \cos x + \cos 3x \sin x\) and \(\sin(3x - x) = \sin 2x = \sin 3x \cos x - \cos 3x \sin x\). Adding these gives \(\sin 4x + \sin 2x = 2 \sin 3x \cos x\), hence \(\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\).

(ii) Integrate \(\frac{1}{2}(\sin 4x + \sin 2x)\) from \(x = 0\) to \(x = \frac{1}{2}\pi\):

\(\int_0^{\frac{1}{2}\pi} \frac{1}{2}(\sin 4x + \sin 2x) \, dx = \frac{1}{2} \left[ -\frac{1}{4} \cos 4x - \frac{1}{2} \cos 2x \right]_0^{\frac{1}{2}\pi}\)

\(= \frac{1}{2} \left( \left(-\frac{1}{4} \cos 2\pi - \frac{1}{2} \cos \pi \right) - \left(-\frac{1}{4} \cos 0 - \frac{1}{2} \cos 0 \right) \right)\)

\(= \frac{1}{2} \left( 0 - \left(-\frac{1}{4} - \frac{1}{2} \right) \right) = \frac{1}{2} \left( \frac{3}{4} \right) = \frac{3}{8}\)

Thus, the area \(R = \frac{9}{16}\).

(iii) Differentiate \(y = \sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\):

\(\frac{dy}{dx} = 2 \cos 4x + \cos 2x\).

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

\(2 \cos 4x + \cos 2x = 0\)

Using double angle formulas, solve for \(\cos 2x\):

\(4 \cos^2 2x + \cos 2x - 2 = 0\)

\(\cos 2x = \frac{-1 \pm \sqrt{33}}{8}\)

Calculate \(x\) for \(\cos 2x = 0.593\), giving \(x = 1.29\).

(i) To find the \(x\)-coordinate of the maximum point \(M\), we first find the derivative of \(y = \frac{x^2}{1 + x^3}\) using the quotient rule. The derivative is:

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

Simplifying, we get:

\(\frac{2x + 2x^4 - 3x^4}{(1 + x^3)^2} = \frac{2x - x^4}{(1 + x^3)^2}\)

Setting the derivative equal to zero to find critical points:

\(2x - x^4 = 0\)

\(x(2 - x^3) = 0\)

\(x = 0\) or \(x^3 = 2\)

Since \(x > 0\), we have \(x = \sqrt[3]{2}\).

(ii) To find \(p\) such that the area of \(R\) is 1, we calculate the definite integral of \(y = \frac{x^2}{1 + x^3}\) from 1 to \(p\):

\(\int_1^p \frac{x^2}{1 + x^3} \, dx = \left[ \frac{1}{3} \ln(1 + x^3) \right]_1^p\)

\(= \frac{1}{3} \ln(1 + p^3) - \frac{1}{3} \ln(2)\)

Setting the area equal to 1:

\(\frac{1}{3} \ln(1 + p^3) - \frac{1}{3} \ln(2) = 1\)

\(\ln(1 + p^3) = 3 + \ln(2)\)

\(1 + p^3 = e^{3 + \ln(2)}\)

\(p^3 = e^{3 + \ln(2)} - 1\)

Solving for \(p\), we find \(p \approx 3.40\) to 3 significant figures.

(i) Differentiate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the maximum point. The derivative is:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x\).

Set the derivative to zero:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x = 0\).

Solve for \(\cos \frac{1}{2}x = \frac{1}{2}\).

This gives \(\alpha = \frac{2}{3}\pi\).

(ii) Integrate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the area:

\(\int_0^{\frac{2}{3}\pi} \left( 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x \right) \, dx\).

The integral is:

\(-16 \cos \frac{1}{2}x + 2 \ln \cos \frac{1}{2}x\).

Evaluate from \(0\) to \(\frac{2}{3}\pi\):

\(8 + 2 \ln \frac{1}{2}\).

(i) To find the maximum point \(M\), we need to find the derivative of the function \(y = e^{-x} - e^{-2x}\) and set it to zero.

The derivative is \(\frac{dy}{dx} = -e^{-x} - (-2)e^{-2x} = -e^{-x} + 2e^{-2x}\).

Setting the derivative to zero gives:

\(-e^{-x} + 2e^{-2x} = 0\)

\(e^{-x} = 2e^{-2x}\)

\(e^{x} = 2\)

\(x = \ln 2\)

Thus, \(p = \ln 2\).

(ii) To find the area of the shaded region, we need to integrate the function from \(x = 0\) to \(x = p\).

The indefinite integral of \(y = e^{-x} - e^{-2x}\) is:

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

Evaluating from \(x = 0\) to \(x = \ln 2\):

\(\left[-e^{-x} + \frac{1}{2}e^{-2x}\right]_0^{\ln 2} = \left(-e^{-\ln 2} + \frac{1}{2}e^{-2\ln 2}\right) - \left(-e^{0} + \frac{1}{2}e^{0}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{2}\left(\frac{1}{4}\right)\right) - \left(-1 + \frac{1}{2}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{8}\right) - \left(-\frac{1}{2}\right)\)

\(= -\frac{1}{2} + \frac{1}{8} + \frac{1}{2}\)

\(= \frac{1}{8}\)

Thus, the area is \(\frac{1}{8}\).

(i) To find the \(x\)-coordinate of \(M\), we differentiate \(y = \frac{x}{x^2 + 1}\) using the quotient rule: \(\frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}\).

Setting the derivative to zero gives \(1 - x^2 = 0\), so \(x = 1\).

(ii) The area \(R\) is given by the integral \(\int_0^p \frac{x}{x^2 + 1} \, dx\). Using substitution \(u = x^2 + 1\), \(du = 2x \, dx\), the integral becomes \(\frac{1}{2} \int_1^{p^2 + 1} \frac{1}{u} \, du = \frac{1}{2} [\ln(u)]_1^{p^2 + 1} = \frac{1}{2} \ln(p^2 + 1)\).

(iii) To find \(p\) such that the area is 1, solve \(\frac{1}{2} \ln(p^2 + 1) = 1\). This gives \(\ln(p^2 + 1) = 2\), so \(p^2 + 1 = e^2\). Therefore, \(p^2 = e^2 - 1\) and \(p = \sqrt{e^2 - 1} \approx 2.53\).