For part (i), a regular hexagon inscribed in a circle of radius $r$ can be divided into 6 equilateral triangles, each with side $r$. The area of one equilateral triangle is $\frac{\sqrt{3}}{4}r^2$, making the hexagon's total area $6 \times \frac{\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{2}r^2$. Dividing this by the circle's area $\pi r^2$ yields the ratio $\frac{3\sqrt{3}}{2\pi}$.

For part (ii), Question 8 involves an equilateral triangle inscribed in a circle of radius $r$, which has an area of $\frac{3\sqrt{3}}{4}r^2$. Since the hexagon's area is exactly twice this, its ratio to the circle's area is also exactly twice.