(i) The tangents AT and BT are equal in length, and each is equal to \(r \tan x\). The arc length of AB is \(2xr\). The perimeter of the shaded region is \(2r \tan x + 2xr\). This is equal to the circumference of the circle, \(2\pi r\). Therefore, \(2r \tan x + 2xr = 2\pi r\). Dividing by \(2r\), we get \(\tan x + x = \pi\), which rearranges to \(\tan x = \pi - x\).

(ii) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.3 \approx 3.6021\). For \(x = 1\), \(\tan 1 - (\pi - 1) \approx 1.5574 - 2.1416 = -0.5842\). For \(x = 1.3\), \(\tan 1.3 - (\pi - 1.3) \approx 3.6021 - 1.8416 = 1.7605\). The sign change indicates a root between 1 and 1.3.

(iii) Using the iterative formula \(x_{n+1} = \arctan(\pi - x_n)\):

Start with \(x_0 = 1\).

\(x_1 = \arctan(\pi - 1) \approx \arctan(2.1416) \approx 1.1366\).

\(x_2 = \arctan(\pi - 1.1366) \approx \arctan(2.0044) \approx 1.1071\).

\(x_3 = \arctan(\pi - 1.1071) \approx \arctan(2.0345) \approx 1.1102\).

\(x_4 = \arctan(\pi - 1.1102) \approx \arctan(2.0314) \approx 1.1098\).

The root correct to 2 decimal places is 1.11.