(i) Differentiate \(\frac{1-x}{1+x}\) using the quotient rule:

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

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(y = \sqrt{u}\) where \(u = \frac{1-x}{1+x}\).

\(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\).

\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2} = \frac{-1}{(1+x)\sqrt{\frac{1-x}{1+x}}}\).

The gradient of the normal is the negative reciprocal of \(\frac{dy}{dx}\):

\(\text{Gradient of normal} = (1+x)\sqrt{1-x^2}\).

(ii) Differentiate the gradient of the normal \((1+x)\sqrt{1-x^2}\) using the product rule:

Let \(u = 1+x\) and \(v = \sqrt{1-x^2}\).

\(\frac{du}{dx} = 1\), \(\frac{dv}{dx} = \frac{-x}{\sqrt{1-x^2}}\).

\(\frac{d}{dx}((1+x)\sqrt{1-x^2}) = u \frac{dv}{dx} + v \frac{du}{dx} = (1+x)\left(\frac{-x}{\sqrt{1-x^2}}\right) + \sqrt{1-x^2}\).

Set \(\frac{d}{dx}((1+x)\sqrt{1-x^2}) = 0\) and solve for \(x\):

\((1+x)\left(\frac{-x}{\sqrt{1-x^2}}\right) + \sqrt{1-x^2} = 0\).

\(-x(1+x) + (1-x^2) = 0\).

\(-x - x^2 + 1 - x^2 = 0\).

\(1 - x - 2x^2 = 0\).

Solving gives \(x = \frac{1}{2}\).

(i) Since \((x + 2)\) is a factor of \(p(x)\), substituting \(x = -2\) into \(p(x)\) gives:

\(-16 + 4a - 2b - 4 = 0\)

\(4a - 2b = 20\) (Equation 1)

Differentiating \(p(x)\), we get \(p'(x) = 6x^2 + 2ax + b\).

Since \((x + 2)\) is a factor of \(p'(x)\), substituting \(x = -2\) into \(p'(x)\) gives:

\(24 - 4a + b = 0\)

\(-4a + b = -24\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 2: \(b = 4a - 24\)

Substitute into Equation 1:

\(4a - 2(4a - 24) = 20\)

\(4a - 8a + 48 = 20\)

\(-4a = -28\)

\(a = 7\)

Substitute \(a = 7\) into \(b = 4a - 24\):

\(b = 4(7) - 24 = 4\)

Thus, \(a = 7\) and \(b = 4\).

(ii) With \(a = 7\) and \(b = 4\), \(p(x) = 2x^3 + 7x^2 + 4x - 4\).

Since \((x + 2)\) is a factor, \((x + 2)^2\) is also a factor.

Attempt division by \((x + 2)^2\):

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

Thus, the factorisation is \((x + 2)^2(2x - 1)\).

To find the points where the gradient of the tangent is 8, we first need to find the derivative of \(y = \frac{x^2}{1 - 3x}\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x^2\) and \(v = 1 - 3x\).

\(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = -3\).

Thus, the derivative \(\frac{dy}{dx} = \frac{(1 - 3x)(2x) - x^2(-3)}{(1 - 3x)^2} = \frac{2x - 6x^2 + 3x^2}{(1 - 3x)^2} = \frac{2x - 3x^2}{(1 - 3x)^2}\).

Set \(\frac{2x - 3x^2}{(1 - 3x)^2} = 8\) and solve for \(x\).

\(2x - 3x^2 = 8(1 - 3x)^2\)

\(2x - 3x^2 = 8(1 - 6x + 9x^2)\)

\(2x - 3x^2 = 8 - 48x + 72x^2\)

\(75x^2 - 50x + 8 = 0\)

Factor the quadratic: \((15x - 4)(5x - 2) = 0\)

Solutions for \(x\) are \(x = \frac{2}{5}\) and \(x = \frac{4}{15}\).

Substitute back to find \(y\):

For \(x = \frac{2}{5}\), \(y = \frac{(\frac{2}{5})^2}{1 - 3(\frac{2}{5})} = \frac{\frac{4}{25}}{1 - \frac{6}{5}} = \frac{\frac{4}{25}}{-\frac{1}{5}} = -\frac{4}{5}\).

For \(x = \frac{4}{15}\), \(y = \frac{(\frac{4}{15})^2}{1 - 3(\frac{4}{15})} = \frac{\frac{16}{225}}{1 - \frac{12}{15}} = \frac{\frac{16}{225}}{\frac{1}{5}} = \frac{16}{45}\).

Thus, the points are \(\left( \frac{2}{5}, -\frac{4}{5} \right)\) and \(\left( \frac{4}{15}, \frac{16}{45} \right)\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}[u v] = u'v + uv'\).

Let \(u = \sin(x + \frac{1}{3}\pi)\) and \(v = \cos x\).

Then \(u' = \cos(x + \frac{1}{3}\pi)\) and \(v' = -\sin x\).

Thus, \(\frac{dy}{dx} = \cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x\).

(ii) At stationary points, \(\frac{dy}{dx} = 0\).

Using the identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\), we have:

\(\cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x = \cos(2x + \frac{1}{3}\pi)\).

Therefore, \(\cos(2x + \frac{1}{3}\pi) = 0\).

(iii) Solve \(\cos(2x + \frac{1}{3}\pi) = 0\).

This implies \(2x + \frac{1}{3}\pi = \frac{\pi}{2} + n\pi\), where \(n\) is an integer.

For \(n = 0\), \(2x + \frac{1}{3}\pi = \frac{\pi}{2}\).

\(2x = \frac{\pi}{2} - \frac{1}{3}\pi = \frac{1}{6}\pi\).

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

For \(n = 1\), \(2x + \frac{1}{3}\pi = \frac{3\pi}{2}\).

\(2x = \frac{3\pi}{2} - \frac{1}{3}\pi = \frac{7}{6}\pi\).

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

Thus, the exact \(x\)-coordinates of the stationary points are \(x = \frac{1}{12}\pi\) and \(x = \frac{7}{12}\pi\).

(i) Let \(y = \ln(1 + \sin 2x)\). Using the chain rule, \(\frac{dy}{dx} = \frac{1}{1 + \sin 2x} \cdot \frac{d}{dx}(1 + \sin 2x)\).

The derivative of \(1 + \sin 2x\) is \(2 \cos 2x\). Therefore, \(\frac{dy}{dx} = \frac{2 \cos 2x}{1 + \sin 2x}\).

(ii) Let \(y = \frac{\tan x}{x}\). Using the quotient rule, \(\frac{dy}{dx} = \frac{x \cdot \sec^2 x - \tan x \cdot 1}{x^2}\).

This simplifies to \(\frac{x \sec^2 x - \tan x}{x^2}\).