Answer: (i) \(\theta=\dfrac{\pi}{3}\), \(r=6\text{ cm}\). (ii) \(10\pi+6\text{ cm}\). (iii) \(30\pi+9\sqrt3\text{ cm}^2\).

(i) The minor arc has length \(2\pi\) and the major arc is \(5\) times as long, so the major arc has length

\(5\times 2\pi=10\pi.\)

Hence the whole circumference is

\(2\pi+10\pi=12\pi.\)

Since the circumference is also \(2\pi r\),

\(2\pi r=12\pi \quad\Rightarrow\quad r=6.\)

Now the minor arc length is \(r\theta\), so

\(6\theta=2\pi \quad\Rightarrow\quad \theta=\frac{\pi}{3}.\)

(ii) The perimeter of the shaded major segment is the major arc plus the chord \(AB\).

The major arc length is \(10\pi\).

The chord length is

\(AB=2r\sin\frac{\theta}{2}=2(6)\sin\frac{\pi}{6}=12\cdot\frac12=6.\)

So the exact perimeter is

\(\boxed{10\pi+6}.\)

(iii) The shaded region is the major segment, so its area is the area of the major sector plus the area of triangle \(AOB\).

The angle of the major sector is

\(2\pi-\frac{\pi}{3}=\frac{5\pi}{3}.\)

Hence the area of the major sector is

\(\frac12 r^2\cdot\frac{5\pi}{3} = \frac12\cdot 36\cdot \frac{5\pi}{3} = 30\pi.\)

The area of triangle \(AOB\) is

\(\frac12 r^2\sin\theta = \frac12\cdot 36\cdot \sin\frac{\pi}{3} = 18\cdot \frac{\sqrt3}{2} = 9\sqrt3.\)

Therefore the area of the shaded segment is

\(\boxed{30\pi+9\sqrt3}.\)