(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(x = 2t + \sin 2t\) gives \(\frac{dx}{dt} = 2 + 2\cos 2t\).

\(y = 1 - 2\cos 2t\) gives \(\frac{dy}{dt} = 4\sin 2t\).

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4\sin 2t}{2 + 2\cos 2t}\).

Using the double angle formula \(\cos 2t = 1 - 2\sin^2 t\), we simplify:

\(\frac{4\sin 2t}{2 + 2\cos 2t} = \frac{4\sin 2t}{2(1 + \cos 2t)} = \frac{4\sin 2t}{2(2\cos^2 t)} = \frac{2\sin 2t}{2\cos^2 t} = 2\tan t\).

Thus, \(\frac{dy}{dx} = 2\tan t\).

(ii) The gradient of the normal is 2, so the gradient of the tangent is \(-\frac{1}{2}\).

Set \(2\tan t = -\frac{1}{2}\), giving \(\tan t = -\frac{1}{4}\).

Thus, \(t = \arctan\left(-\frac{1}{4}\right)\).

Substitute \(t\) back into \(x = 2t + \sin 2t\) to find \(x\).

\(x = 2\left(\arctan\left(-\frac{1}{4}\right)\right) + \sin\left(2\arctan\left(-\frac{1}{4}\right)\right)\).

Calculating gives \(x \approx -0.961\).

(i) First, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):

\(\frac{dy}{dt} = 4 + \frac{2}{2t - 1}\)

\(\frac{dx}{dt} = 2t\)

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

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 + \frac{2}{2t - 1}}{2t}\)

Simplify to obtain:

\(\frac{dy}{dx} = \frac{8t - 2}{2t(2t - 1)}\)

(ii) To find the equation of the normal, first find the gradient of the tangent at \(t = 1\):

Substitute \(t = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{8(1) - 2}{2(1)(2(1) - 1)} = \frac{6}{2} = 3\)

The gradient of the normal is the negative reciprocal of the tangent's gradient:

\(m_{\text{normal}} = -\frac{1}{3}\)

Find the point on the curve at \(t = 1\):

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

\(y = 4(1) + \ln(2(1) - 1) = 4 + \ln(1) = 4\)

Using the point-slope form of the line equation:

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

Rearrange to the form \(ax + by + c = 0\):

\(3(y - 4) = -(x - 2)\)

\(3y - 12 = -x + 2\)

\(x + 3y - 14 = 0\)

(i) Differentiate \(x = \ln \cos \theta\) with respect to \(\theta\):

\(\frac{dx}{d\theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta\).

Differentiate \(y = 3\theta - \tan \theta\) with respect to \(\theta\):

\(\frac{dy}{d\theta} = 3 - \sec^2 \theta\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{3 - \sec^2 \theta}{-\tan \theta}\).

Simplify to obtain:

\(\frac{dy}{dx} = \frac{\tan^2 \theta - 2}{\tan \theta}\).

(ii) The gradient of the normal is \(-1\), so the gradient of the tangent is \(1\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{\tan^2 \theta - 2}{\tan \theta} = 1\).

Solve the equation:

\(\tan^2 \theta - \tan \theta - 2 = 0\).

Factorize to find \(\tan \theta = 1\) or \(\tan \theta = -2\).

For \(\tan \theta = 1\), \(\theta = \frac{\pi}{4}\).

Substitute \(\theta = \frac{\pi}{4}\) into \(y = 3\theta - \tan \theta\):

\(y = 3 \times \frac{\pi}{4} - 1 = \frac{3\pi}{4} - 1\).

Thus, \(y = -1\).

(i) Differentiate \(x = t + \cos t\) with respect to \(t\):

\(\frac{dx}{dt} = 1 - \sin t\).

Differentiate \(y = \ln(1 + \sin t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{\cos t}{1 + \sin t}\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\frac{\cos t}{1 + \sin t}}{1 - \sin t} = \frac{\cos t}{(1 + \sin t)(1 - \sin t)} = \frac{\cos t}{1 - \sin^2 t} = \frac{\cos t}{\cos^2 t} = \sec t\).

(ii) Set \(\frac{dy}{dx} = 3\):

\(\sec t = 3 \Rightarrow \cos t = \frac{1}{3}\).

Find \(t\) using \(t = \cos^{-1}\left(\frac{1}{3}\right)\).

Calculate \(x = t + \cos t\) for each \(t\):

For \(t = \cos^{-1}\left(\frac{1}{3}\right)\), \(x = 1.56\).

For \(t = -\cos^{-1}\left(\frac{1}{3}\right)\), \(x = -0.898\).

(i) Differentiate \(x = a \cos^4 t\) and \(y = a \sin^4 t\) with respect to \(t\):

\(\frac{dx}{dt} = -4a \cos^3 t \sin t\)

\(\frac{dy}{dt} = 4a \sin^3 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4a \sin^3 t \cos t}{-4a \cos^3 t \sin t} = -\tan t\).

(ii) The equation of the tangent is given by:

\(\frac{y - a \sin^4 t}{x - a \cos^4 t} = -\tan t\)

Rearranging gives:

\(x \sin^2 t + y \cos^2 t = a \sin^2 t \cos^2 t\).

(iii) For the tangent meeting the x-axis at \(P\) and the y-axis at \(Q\):

At \(P\), \(y = 0\), so \(x = a \cos^2 t\).

At \(Q\), \(x = 0\), so \(y = a \sin^2 t\).

Thus, \(OP = a \cos^2 t\) and \(OQ = a \sin^2 t\).

Therefore, \(OP + OQ = a \cos^2 t + a \sin^2 t = a(\cos^2 t + \sin^2 t) = a\).