(i) The perimeter of R₁ is given by the sum of the two radii and the arc length: \(r + r + r\theta = 2r + r\theta\).

The length of the major arc AB is \(2\pi r - r\theta\).

Equating the two expressions: \(2r + r\theta = 2\pi r - r\theta\).

Solving for \(\theta\):

\(2r + r\theta = 2\pi r - r\theta\)

\(2r + r\theta + r\theta = 2\pi r\)

\(2r + 2r\theta = 2\pi r\)

\(2r(1 + \theta) = 2\pi r\)

\(1 + \theta = \pi\)

\(\theta = \pi - 1\)

(ii) The area of region R₁ is given by \(\frac{1}{2}r^2\theta\).

Given \(\theta = \pi - 1\) and area of R₁ is 30 cm²:

\(\frac{1}{2}r^2(\pi - 1) = 30\)

\(r^2 = \frac{60}{\pi - 1}\)

\(r = \sqrt{\frac{60}{\pi - 1}} \approx 5.29\)

The area of region R₂ is the area of the circle minus the area of R₁:

\(R_2 = \frac{1}{2}r^2(\pi + 1)\)

\(R_2 = \frac{1}{2} \times \frac{60}{\pi - 1} \times (\pi + 1)\)

\(R_2 \approx 58.0 \text{ cm}^2\)