(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 \(x \approx -0.961\).