To find the stationary points, we need to differentiate the function and set the derivative equal to zero.

The given function is \(y = 3 \cos 2x + 7 \sin x + 2\).

Differentiate with respect to \(x\):

\(\frac{dy}{dx} = -6 \sin 2x + 7 \cos x\).

Using the identity \(\sin 2x = 2 \sin x \cos x\), rewrite the derivative:

\(\frac{dy}{dx} = -6 (2 \sin x \cos x) + 7 \cos x\).

\(\frac{dy}{dx} = -12 \sin x \cos x + 7 \cos x\).

Factor out \(\cos x\):

\(\frac{dy}{dx} = \cos x (-12 \sin x + 7)\).

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

\(\cos x (-12 \sin x + 7) = 0\).

This gives two equations:

1. \(\cos x = 0\)

2. \(-12 \sin x + 7 = 0\)

For \(\cos x = 0\), \(x = \frac{\pi}{2} = 1.57\).

For \(-12 \sin x + 7 = 0\):

\(\sin x = \frac{7}{12}\).

Solving for \(x\) in the interval \(0 \leq x \leq \pi\):

\(x = \arcsin \left( \frac{7}{12} \right) \approx 0.623\).

\(x = \pi - \arcsin \left( \frac{7}{12} \right) \approx 2.52\).

Thus, the \(x\)-coordinates of the stationary points are \(0.623, 1.57, 2.52\).

To find the stationary point, we need to differentiate \(y = \sin x \sin 2x\) and set the derivative to zero.

Using the product rule, \(y' = \cos x \sin 2x + 2 \sin x \cos 2x\).

Using the double angle formula \(\sin 2x = 2 \sin x \cos x\), we can express the derivative in terms of \(\sin x\) and \(\cos x\).

Equating the derivative to zero, we obtain \(3 \sin^2 x = 2\), which simplifies to \(\tan^2 x = 2\).

Solving for \(x\), we find \(x = 0.955\) (to 3 significant figures).

To find the gradient of the curve, we need to differentiate \(y = \frac{1+x}{1+2x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = 1+x\) and \(v = 1+2x\).

First, find \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = 2\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(1+2x)(1) - (1+x)(2)}{(1+2x)^2}\)

Simplify the numerator:

\((1+2x) - 2(1+x) = 1 + 2x - 2 - 2x = -1\)

Thus, \(\frac{dy}{dx} = \frac{-1}{(1+2x)^2}\).

Since \((1+2x)^2 > 0\) for \(x > -\frac{1}{2}\), \(\frac{-1}{(1+2x)^2} < 0\).

Therefore, the gradient of the curve is always negative for \(x > -\frac{1}{2}\).

(i) To find the stationary point, we need to find the derivative \(y'\) and set it to zero. Using the quotient rule for \(y = \frac{e^{2x}}{x^3}\), we have:

\(y' = \frac{(2e^{2x})(x^3) - (e^{2x})(3x^2)}{(x^3)^2} = \frac{2e^{2x}x^3 - 3e^{2x}x^2}{x^6}\)

Simplifying, \(y' = \frac{e^{2x}(2x^3 - 3x^2)}{x^6} = \frac{e^{2x}x^2(2x - 3)}{x^6} = \frac{e^{2x}(2x - 3)}{x^4}\)

Setting \(y' = 0\), we solve \(2x - 3 = 0\), giving \(x = \frac{3}{2}\).

(ii) To determine the nature of the stationary point, we test the sign of \(y'\) around \(x = \frac{3}{2}\). For \(x < \frac{3}{2}\), \(2x - 3 < 0\), so \(y' < 0\). For \(x > \frac{3}{2}\), \(2x - 3 > 0\), so \(y' > 0\). This change from negative to positive indicates a minimum point.

(i) To find the stationary points, we first find the derivative of \(y = 3 \sin x + 4 \cos^3 x\).

The derivative is \(\frac{dy}{dx} = 3 \cos x - 12 \cos^2 x \sin x\).

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

\(3 \cos x - 12 \cos^2 x \sin x = 0\)

Factor out \(\cos x\):

\(\cos x (3 - 12 \cos x \sin x) = 0\)

This gives \(\cos x = 0\) or \(3 - 12 \cos x \sin x = 0\).

For \(\cos x = 0\), \(x = \frac{1}{2}\pi\).

For \(3 - 12 \cos x \sin x = 0\), solve for \(\sin 2x = \frac{1}{2}\):

\(2x = \frac{\pi}{6}, \frac{5\pi}{6}\)

Thus, \(x = \frac{1}{12}\pi, \frac{5}{12}\pi\).

Therefore, the \(x\)-coordinates are \(x = \frac{1}{12}\pi, \frac{5}{12}\pi, \frac{1}{2}\pi\).

(ii) To determine the nature of the stationary point at \(x = \frac{1}{12}\pi\), we check the second derivative or use the first derivative test.

Since the mark scheme indicates a maximum at \(x = \frac{1}{12}\pi\), this is a maximum point.