(a) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = \tan x\), then \(y = \sqrt{u}\). The derivative \(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\) and \(\frac{du}{dx} = \sec^2 x\). Therefore, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x = \frac{1 + \tan^2 x}{2\sqrt{\tan x}}\).

Verify at \(x = \frac{1}{4}\pi\): \(\tan \frac{1}{4}\pi = 1\), so \(\frac{dy}{dx} = \frac{1 + 1^2}{2\sqrt{1}} = 1\).

(b) Substitute \(t = \tan a\) into the equation from part (a): \(\frac{1 + t^2}{2\sqrt{t}} = 1\). Rearrange to get \((1 + t^2)^2 = 4t\), leading to \(t^4 + 2t^2 - 4t + 1 = 0\). Divide by \(t - 1\) to obtain \(t^3 + t^2 + 3t - 1 = 0\).

(c) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{1}{3} (1 - \tan^2 a_n - \tan^3 a_n) \right)\). Start with an initial guess, e.g., \(a_0 = 0.3\), and iterate: \(a_1 = 0.2854\), \(a_2 = 0.2894\), \(a_3 = 0.2883\), etc. Continue until \(a\) stabilizes to 2 decimal places: \(a = 0.29\).