(a) The length of the tangent from a point to a circle is given by \(AT = r \tan x\) and \(BT = r \tan x\). The area of the sector \(AOB\) is \(\frac{1}{2} r^2 (2x) = r^2 x\). The area of the triangle \(AOT\) is \(\frac{1}{2} r \cdot r \tan x = \frac{1}{2} r^2 \tan x\). The area of the shaded region is the area of the circle minus the area of the sector, which is \(\pi r^2 - r^2 x\). Equating the area of the shaded region to the area of the circle, we have:

\(\pi r^2 - r^2 x = r^2 \tan x\)

\(\pi - x = \tan x\)

Thus, \(x\) satisfies \(\tan x = \pi + x\).

(b) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.4 \approx 5.7979\). Since \(\pi + 1 \approx 4.1416\) and \(\pi + 1.4 \approx 4.5416\), \(\tan x = \pi + x\) changes sign between 1 and 1.4, confirming a root in this interval.

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

\(x_1 = \arctan(\pi) \approx 1.2626\)

\(x_2 = \arctan(\pi + 1.2626) \approx 1.3511\)

\(x_3 = \arctan(\pi + 1.3511) \approx 1.3521\)

\(x_4 = \arctan(\pi + 1.3521) \approx 1.3521\)

The root correct to 2 decimal places is 1.35.