Answer: (i) The curve is a single loop of \(r=\cos 2\theta\), symmetric about the \(x\)-axis, from \(\theta=-\frac{\pi}{4}\) to \(\theta=\frac{\pi}{4}\), with maximum radius \(1\) at \(\theta=0\).

(ii) The enclosed area is \(\frac{\pi}{8}\).

(iii) A Cartesian equation is \(\left(x^2+y^2\right)^{3/2}=x^2-y^2\).

(i) The curve meets the pole when

\(\cos 2\theta=0\),

so \(2\theta=\pm\frac{\pi}{2}\), and therefore \(\theta=\pm\frac{\pi}{4}\).

At \(\theta=0\), \(r=1\), so the curve reaches \((1,0)\). On \(-\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}\), \(r\geq0\), giving one loop symmetric about the \(x\)-axis.

(ii) The polar area is

\(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4} r^2\,d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta.\)

Using \(\cos^2 2\theta=\frac{1}{2}(1+\cos4\theta)\),

\(A=\frac{1}{4}\left[\theta+\frac{1}{4}\sin4\theta\right]_{-\pi/4}^{\pi/4}.\)

Since \(\sin\pi=\sin(-\pi)=0\),

\(A=\frac{1}{4}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi}{8}.\)

So the area enclosed by \(C\) is \(\boxed{\frac{\pi}{8}}\).

(iii) Since

\(\cos 2\theta=\cos^2\theta-\sin^2\theta=\frac{x^2}{r^2}-\frac{y^2}{r^2}=\frac{x^2-y^2}{r^2},\)

the equation \(r=\cos2\theta\) gives

\(r=\frac{x^2-y^2}{r^2}.\)

Hence

\(r^3=x^2-y^2.\)

Using \(r^2=x^2+y^2\), we get

\(\boxed{\left(x^2+y^2\right)^{3/2}=x^2-y^2}.\)