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

\(x = \tan \theta \Rightarrow \frac{dx}{d\theta} = \sec^2 \theta\).

\(y = 2 \cos^2 \theta \sin \theta\).

Using the product rule, \(\frac{dy}{d\theta} = 2 \left(2 \cos \theta (-\sin \theta) \sin \theta + \cos^2 \theta \cos \theta\right) = -4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta\).

Now, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta}{\sec^2 \theta}\).

\(\frac{dy}{dx} = (-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta) \cos^2 \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute:

\(-4 \cos \theta (1 - \cos^2 \theta) + 2 \cos^3 \theta = -4 \cos \theta + 4 \cos^3 \theta + 2 \cos^3 \theta = -4 \cos \theta + 6 \cos^3 \theta\).

Thus, \(\frac{dy}{dx} = 6 \cos^5 \theta - 4 \cos^3 \theta\).

(a) To find \(\frac{dy}{dx}\), we use the chain rule:

\(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\).

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

\(x = \sqrt{t} + 3 \Rightarrow \frac{dx}{dt} = \frac{1}{2\sqrt{t}}\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln t \Rightarrow \frac{dy}{dt} = \frac{1}{t}\).

Thus, \(\frac{dy}{dx} = \frac{1}{t} \times 2\sqrt{t} = \frac{2}{\sqrt{t}}\).

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

Set \(\frac{dy}{dx} = \frac{1}{2}\):

\(\frac{2}{\sqrt{t}} = \frac{1}{2} \Rightarrow \sqrt{t} = 4 \Rightarrow t = 16\).

Substitute \(t = 16\) into the parametric equations:

\(x = \sqrt{16} + 3 = 7\),

\(y = \ln 16\).

Therefore, the coordinates are \((7, \ln 16)\).

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

\(x = \frac{1}{\cos^3 t} \Rightarrow \frac{dx}{dt} = \frac{3 \sin t}{\cos^4 t}\)

\(y = \tan^3 t \Rightarrow \frac{dy}{dt} = 3 \tan^2 t \sec^2 t\)

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \tan^2 t \sec^2 t}{\frac{3 \sin t}{\cos^4 t}} = \sin t\)

(ii) The equation of the tangent to the curve at a point is given by \(y - y_1 = m(x - x_1)\), where \(m = \frac{dy}{dx}\).

At parameter \(t\), \(x_1 = \frac{1}{\cos^3 t}\) and \(y_1 = \tan^3 t\).

Substituting, \(y - \tan^3 t = \sin t \left(x - \frac{1}{\cos^3 t}\right)\)

Simplifying, \(y = x \sin t - \tan t\)

(i) Differentiate \(x = t - \tan t\) with respect to \(t\):

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

Differentiate \(y = \ln(\cos t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{-\sin t}{\cos t} = -\tan t\).

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):

\(\frac{dy}{dx} = \frac{-\tan t}{1 - \sec^2 t} = \cot t\).

(ii) To find the \(x\)-coordinate where the gradient is 2, set \(\frac{dy}{dx} = 2\):

\(\cot t = 2 \Rightarrow t = \arctan\left(\frac{1}{2}\right)\).

Substitute \(t = \arctan\left(\frac{1}{2}\right)\) into \(x = t - \tan t\):

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

Calculate \(x \approx -0.0364\) to 3 significant figures.

To find the gradient of the curve at the point where it crosses the y-axis, we first need to determine the value of \(t\) when \(x = 0\).

Since \(x = \\ln(2t + 3)\), setting \(x = 0\) gives:

\(\\ln(2t + 3) = 0\)

\(2t + 3 = 1\)

\(2t = -2\)

\(t = -1\)

Now, we find \(\frac{dy}{dx}\) using the chain rule:

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

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

\(x = \\ln(2t + 3)\)

\(\frac{dx}{dt} = \frac{2}{2t + 3}\)

Next, find \(\frac{dy}{dt}\):

\(y = \frac{3t + 2}{2t + 3}\)

Using the quotient rule:

\(\frac{dy}{dt} = \frac{(2t + 3)(3) - (3t + 2)(2)}{(2t + 3)^2}\)

\(= \frac{6t + 9 - (6t + 4)}{(2t + 3)^2}\)

\(= \frac{5}{(2t + 3)^2}\)

Now, substitute \(t = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{5}{(2(-1) + 3)^2}}{\frac{2}{2(-1) + 3}}\)

\(= \frac{\frac{5}{1^2}}{\frac{2}{1}}\)

\(= \frac{5}{2}\)

Thus, the gradient of the curve at the point where it crosses the y-axis is \(\frac{5}{2}\).