Answer: Cartesian equation: \(xy=4\).

Sketch: A branch of the rectangular hyperbola in the first quadrant, approaching both axes as asymptotes.

Exact area: \(2\ln 3\) (equivalently \(\ln 9\)).

Using the polar relations \(x=r\cos\theta\) and \(y=r\sin\theta\), we have

\(r^2=8\operatorname{cosec} 2\theta=\dfrac{8}{\sin 2\theta}=\dfrac{8}{2\sin\theta\cos\theta}=\dfrac{4}{\sin\theta\cos\theta}.\)

Multiplying by \(\sin\theta\cos\theta\) gives

\(r^2\sin\theta\cos\theta=4\).

Since \(r\cos\theta=x\) and \(r\sin\theta=y\), this becomes \(xy=4\).

Hence the cartesian equation of the curve is \(xy=4\).

For the sketch, since \(0\lt \theta\lt \frac{\pi}{2}\), both \(x\) and \(y\) are positive, so the curve lies in the first quadrant only. Also, as \(x\to 0^+\), \(y=4/x\to\infty\), and as \(y\to 0^+\), \(x\to\infty\), so the axes are asymptotes. The curve is therefore the first-quadrant branch of the hyperbola \(xy=4\).

For the area of the sector, use the polar area formula:

\(\text{Area}=\dfrac{1}{2}\int_{\pi/6}^{\pi/3} r^2\,d\theta=\dfrac{1}{2}\int_{\pi/6}^{\pi/3} 8\operatorname{cosec} 2\theta\,d\theta=4\int_{\pi/6}^{\pi/3}\operatorname{cosec} 2\theta\,d\theta.\)

Let \(u=2\theta\), so \(du=2\,d\theta\) and \(d\theta=\dfrac{1}{2}du\). Then

\(\text{Area}=2\int_{\pi/3}^{2\pi/3}\operatorname{cosec} u\,du.\)

Using \(\int \operatorname{cosec} u\,du=\ln\left|\tan\dfrac{u}{2}\right|+C\),

\(\text{Area}=2\left[\ln\left|\tan\dfrac{u}{2}\right|\right]_{\pi/3}^{2\pi/3}=2\left(\ln\tan\dfrac{\pi}{3}-\ln\tan\dfrac{\pi}{6}\right).\)

Since \(\tan\dfrac{\pi}{3}=\sqrt{3}\) and \(\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\),

\(\text{Area}=2\left(\ln\sqrt{3}-\ln\dfrac{1}{\sqrt{3}}\right)=2\ln 3.\)