Answer:

1. The polar equation is \(r=\sin 4\theta\).
2. The curve is a single loop in the first quadrant, starting and ending at the origin. Its line of symmetry is \(\theta=\frac{\pi}{8}\).
3. The area enclosed by \(C\) is \(\frac{\pi}{16}\).
4. The maximum distance from the line \(\theta=\frac{\pi}{2}\) is \(0.93\) to 2 decimal places.

1. Use \(x=r\cos\theta\), \(y=r\sin\theta\) and \(x^2+y^2=r^2\).

   The Cartesian equation is

   \((x^2+y^2)^{5/2}=4xy(x^2-y^2)\).

   Substituting gives

   \(r^5=4(r\cos\theta)(r\sin\theta)(r^2\cos^2\theta-r^2\sin^2\theta)\),

   so

   \(r^5=4r^4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\).

   For \(r e 0\), divide by \(r^4\):

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

   Using \(2\sin\theta\cos\theta=\sin2\theta\) and \(\cos^2\theta-\sin^2\theta=\cos2\theta\),

   \(r=2\sin2\theta\cos2\theta=\sin4\theta\).

   Thus the polar equation is \(r=\sin4\theta\).

2. For \(0\leq\theta\leq\frac{\pi}{4}\), \(r=\sin4\theta\geq0\). Also \(r=0\) at \(\theta=0\) and \(\theta=\frac{\pi}{4}\), and \(r\) is greatest when \(4\theta=\frac{\pi}{2}\), i.e. \(\theta=\frac{\pi}{8}\).

   So \(C\) is one loop in the first quadrant, starting at the origin on the initial line, reaching its furthest point on \(\theta=\frac{\pi}{8}\), and returning to the origin on \(\theta=\frac{\pi}{4}\).

   The line of symmetry is \(\theta=\frac{\pi}{8}\).

3. The area enclosed by a polar curve is

   \(A=\frac12\int_\alpha^\beta r^2\,d\theta\).

   Here \(r=\sin4\theta\), with \(0\leq\theta\leq\frac{\pi}{4}\), so

   \(A=\frac12\int_0^{\pi/4}\sin^2 4\theta\,d\theta\).

   Using \(\sin^2 u=\frac12(1-\cos2u)\),

   \(A=\frac12\int_0^{\pi/4}\frac12(1-\cos8\theta)\,d\theta\)

   \(=\frac14\left[\theta-\frac18\sin8\theta\right]_0^{\pi/4}\).

   Since \(\sin2\pi=0\),

   \(A=\frac14\cdot\frac{\pi}{4}=\frac{\pi}{16}\).

4. The line \(\theta=\frac{\pi}{2}\) is the \(y\)-axis, so the distance of a point on this loop from the line is \(x\), since \(x\geq0\) on \(C\).

   Now \(x=r\cos\theta\). Using the given identity,

   \(x=\sin4\theta\cos\theta\)

   \(=(4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta)\cos\theta\)

   \(=4\sin\theta\cos^4\theta-4\sin^3\theta\cos^2\theta\).

   Differentiate and set equal to zero:

   \(\frac{dx}{d\theta}=4\cos\theta(\cos^4\theta-7\sin^2\theta\cos^2\theta+2\sin^4\theta)=0\).

   In \(0\leq\theta\leq\frac{\pi}{4}\), \(\cos\theta e 0\), so

   \(\cos^4\theta-7\sin^2\theta\cos^2\theta+2\sin^4\theta=0\).

   Dividing by \(\cos^4\theta\),

   \(1-7\tan^2\theta+2\tan^4\theta=0\).

   Let \(u=\tan^2\theta\). Then

   \(2u^2-7u+1=0\),

   so

   \(u=\frac{7\pm\sqrt{41}}{4}\).

   Since \(0\leq\tan^2\theta\leq1\),

   \(\tan^2\theta=\frac{7-\sqrt{41}}{4}\).

   Hence \(\theta\approx0.369\). Substituting into \(x=\sin4\theta\cos\theta\),

   \(x\approx0.929\).

   The endpoints give \(x=0\), so the maximum distance from \(\theta=\frac{\pi}{2}\) is \(0.93\) to 2 decimal places.