(i) Differentiate \(f(x) = (x-2)^2 g(x)\) to get \(f'(x) = 2(x-2)g(x) + (x-2)^2 g'(x)\). Since \((x-2)\) is a common factor, \((x-2)\) is a factor of \(f'(x)\).

(ii) Given \(x^5 + ax^4 + 3x^3 + bx^2 + a\) has a factor \((x-2)^2\), substitute \(x = 2\) to get \(32 + 16a + 24 + 4b + a = 0\), simplifying to \(32 + 17a + 4b = 0\).

Differentiate the polynomial to get \(5x^4 + 4ax^3 + 9x^2 + 2bx\). Substitute \(x = 2\) to get \(80 + 32a + 36 + 4b = 0\), simplifying to \(116 + 32a + 4b = 0\).

Solving the equations \(32 + 17a + 4b = 0\) and \(116 + 32a + 4b = 0\) simultaneously, we find \(a = -4\) and \(b = 3\).

(i) To find the stationary points, we differentiate the function \(y = 10e^{-\frac{1}{2}x} \sin 4x\) using the product rule:

\(\frac{dy}{dx} = -5e^{-\frac{1}{2}x} \sin 4x + 40e^{-\frac{1}{2}x} \cos 4x\).

Setting \(\frac{dy}{dx} = 0\), we solve:

\(\tan 4x = 8\).

This gives \(x = 0.362\) or \(x = 1.147\) (in radians).

(ii) The \(x\)-coordinates of \(T_n\) increase by \(\frac{\pi}{4}\). We solve the inequality:

\(x_1 + (n-1)\frac{\pi}{4} > 25\).

Substituting \(x_1 = 0.362\), we get:

\(n > \frac{4}{\pi}(25 - 0.362) + 1\).

Calculating gives \(n = 33\).