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

\(\frac{dx}{dt} = \frac{d}{dt}(\ln(\tan t)) = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

\(\frac{dy}{dt} = \frac{d}{dt}(\sin^2 t) = 2 \sin t \cos t\).

Then, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \sin t \cos t}{\sec^2 t / \tan t} = 2 \sin^2 t \cos^2 t\).

(ii) To find the equation of the tangent at \(x = 0\), we first find \(t\) when \(x = 0\).

\(x = \ln(\tan t) = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(y = \sin^2(\frac{\pi}{4}) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\).

The gradient \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\) is \(\frac{1}{2}\).

The equation of the tangent is \(y - \frac{1}{2} = \frac{1}{2}(x - 0)\).

Simplifying, \(y = \frac{1}{2}x + \frac{1}{2}\).

To find the gradient \(\frac{dy}{dx}\), we first need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(x = \frac{t}{2t + 3}\), use the quotient rule:

\(\frac{dx}{dt} = \frac{(2t + 3)(1) - t(2)}{(2t + 3)^2} = \frac{3}{(2t + 3)^2}\).

For \(y = e^{-2t}\), differentiate with respect to \(t\):

\(\frac{dy}{dt} = -2e^{-2t}\).

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

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2e^{-2t}}{\frac{3}{(2t + 3)^2}} = -2e^{-2t} \cdot \frac{(2t + 3)^2}{3}\).

Evaluate at \(t = 0\):

\(\frac{dy}{dx} = -2e^{0} \cdot \frac{3^2}{3} = -2 \cdot 3 = -6\).

First, find \(\frac{dy}{d\theta}\):

\(\frac{dy}{d\theta} = 1 - 2\sin \theta\).

Next, find \(\frac{dx}{d\theta}\) using the quotient rule:

\(x = \frac{\cos \theta}{2 - \sin \theta}\).

Let \(u = \cos \theta\) and \(v = 2 - \sin \theta\).

Then \(\frac{du}{d\theta} = -\sin \theta\) and \(\frac{dv}{d\theta} = -\cos \theta\).

Using the quotient rule, \(\frac{dx}{d\theta} = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2}\).

\(\frac{dx}{d\theta} = \frac{(2 - \sin \theta)(-\sin \theta) + \cos \theta \cdot \cos \theta}{(2 - \sin \theta)^2}\).

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + \sin^2 \theta + \cos^2 \theta}{(2 - \sin \theta)^2}\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\),

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}\).

Now, use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\):

\(\frac{dy}{dx} = \frac{1 - 2\sin \theta}{\frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}}\).

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

(i) Differentiate \(x = a \cos^3 t\) and \(y = a \sin^3 t\) with respect to \(t\):

\(\frac{dx}{dt} = -3a \cos^2 t \sin t\)

\(\frac{dy}{dt} = 3a \sin^2 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\tan t}{\cos t} = -\tan t \sec t\).

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

\(\frac{y - a \sin^3 t}{x - a \cos^3 t} = \frac{dy}{dx} = -\tan t \sec t\)

Rearranging gives:

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

(iii) The tangent meets the \(x\)-axis when \(y = 0\):

\(x \sin t = a \sin t \cos t \Rightarrow x = a \cos t\).

The tangent meets the \(y\)-axis when \(x = 0\):

\(y \cos t = a \sin t \cos t \Rightarrow y = a \sin t\).

The length \(XY\) is:

\(\sqrt{(a \cos t)^2 + (a \sin t)^2} = \sqrt{a^2 (\cos^2 t + \sin^2 t)} = a\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = a(2 - 2\cos 2\theta) = 2a(1 - \cos 2\theta)\)

\(\frac{dy}{d\theta} = a(2\sin 2\theta)\)

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

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a\sin 2\theta}{2a(1 - \cos 2\theta)}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{\sin 2\theta}{1 - \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 - \cos 2\theta = 2\sin^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\sin^2 \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)