(i) To find the stationary point, we need to differentiate \(y = \frac{2 - \\sin x}{\\cos x}\).

Rewrite the function as \(y = 2\\sec x - \\tan x\).

The derivative is \(y' = 2\\sec x \\tan x - \\sec^2 x\).

Set the derivative to zero: \(2\\sec x \\tan x - \\sec^2 x = 0\).

Factor to get \(\\sec x (2\\tan x - \\sec x) = 0\).

Since \(\\sec x \neq 0\), solve \(2\\tan x = \\sec x\).

Using \(\\tan x = \\frac{\\sin x}{\\cos x}\) and \(\\sec x = \\frac{1}{\\cos x}\), we have \(2\\frac{\\sin x}{\\cos x} = \\frac{1}{\\cos x}\).

Simplify to \(2\\sin x = 1\), giving \(\\sin x = \\frac{1}{2}\).

In the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), \(x = \frac{1}{6}\pi\).

Substitute back to find \(y\): \(y = \frac{2 - \\sin(\frac{1}{6}\pi)}{\\cos(\frac{1}{6}\pi)} = \sqrt{3}\).

Thus, the coordinates are \(\left( \frac{1}{6}\pi, \sqrt{3} \right)\).

(ii) To determine the nature of the stationary point, consider the second derivative or test values around \(x = \frac{1}{6}\pi\).

Since the mark scheme indicates it is a minimum, we conclude it is a minimum point.