(i) To find the stationary points, we need to differentiate \(y = (1 + x^2) e^{-\frac{1}{2}x}\) and set the derivative to zero. Using the product rule, let \(u = 1 + x^2\) and \(v = e^{-\frac{1}{2}x}\). Then \(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = -\frac{1}{2} e^{-\frac{1}{2}x}\).

The derivative is \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (1 + x^2)(-\frac{1}{2} e^{-\frac{1}{2}x}) + e^{-\frac{1}{2}x}(2x)\).

Simplifying, \(\frac{dy}{dx} = e^{-\frac{1}{2}x}(-\frac{1}{2} - \frac{1}{2}x^2 + 2x)\).

Setting \(\frac{dy}{dx} = 0\), we solve \(-\frac{1}{2} - \frac{1}{2}x^2 + 2x = 0\).

Multiplying through by \(-2\), we get \(x^2 - 4x - 1 = 0\).

Using the quadratic formula, \(x = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{3}\).

(ii) To find the area of \(R\), we integrate \(y = (1 + x^2) e^{-\frac{1}{2}x}\) from \(x = 0\) to \(x = 2\).

Using integration by parts, let \(u = 1 + x^2\) and \(dv = e^{-\frac{1}{2}x} dx\). Then \(du = 2x dx\) and \(v = -2e^{-\frac{1}{2}x}\).

The integral becomes \(\int (1 + x^2) e^{-\frac{1}{2}x} dx = -2(1 + x^2) e^{-\frac{1}{2}x} + \int 4x e^{-\frac{1}{2}x} dx\).

Integrating by parts again for \(\int 4x e^{-\frac{1}{2}x} dx\), we find \(-8xe^{-\frac{1}{2}x} - 2x^2 e^{-\frac{1}{2}x}\).

Evaluating from \(x = 0\) to \(x = 2\), we find the area is \(18 - \frac{42}{e}\).

(i) The derivative of \(y = (\ln x)^2\) is \(\frac{2 \ln x}{x}\). At \(x = e\), the gradient of the tangent is \(\frac{2 \ln e}{e} = \frac{2}{e}\). The gradient of the normal is the negative reciprocal, \(-\frac{e}{2}\). The equation of the normal at \(P(e, 1)\) is \(y - 1 = -\frac{e}{2}(x - e)\). Setting \(y = 0\) to find \(Q\), we solve \(0 - 1 = -\frac{e}{2}(x - e)\), giving \(x = e + \frac{2}{e}\).

(ii) To show \(\int \ln x \, dx = x \ln x - x + c\), use integration by parts: let \(u = \ln x\) and \(dv = dx\), then \(du = \frac{1}{x} dx\) and \(v = x\). Thus, \(\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + c\).

(iii) The area under the curve from \(x = 1\) to \(x = e\) is \(\int_1^e (\ln x)^2 \, dx\). Using integration by parts, let \(u = (\ln x)^2\) and \(dv = dx\), then \(du = \frac{2 \ln x}{x} dx\) and \(v = x\). Thus, \(\int (\ln x)^2 \, dx = x(\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} \, dx = x(\ln x)^2 - 2 \int \ln x \, dx\). From part (ii), \(\int \ln x \, dx = x \ln x - x\), so \(\int (\ln x)^2 \, dx = x(\ln x)^2 - 2(x \ln x - x) = x(\ln x)^2 - 2x \ln x + 2x\). Evaluating from \(x = 1\) to \(x = e\), we find the area is \(e - 2\). The area of the triangle formed by the normal is \(\frac{1}{2} \times \frac{2}{e} \times 1 = \frac{1}{e}\). The total area is \(e - 2 + \frac{1}{e}\).

(i) To find the \(x\)-coordinate of the maximum point \(M\), we need to find the derivative of \(y = (2x - x^2)e^{\frac{1}{2}x}\) and set it to zero.

Using the product rule, the derivative is:

\(\frac{d}{dx}[(2x - x^2)e^{\frac{1}{2}x}] = (2 - 2x)e^{\frac{1}{2}x} + \frac{1}{2}(2x - x^2)e^{\frac{1}{2}x}\)

Setting the derivative to zero:

\((2 - 2x)e^{\frac{1}{2}x} + \frac{1}{2}(2x - x^2)e^{\frac{1}{2}x} = 0\)

Simplifying and solving for \(x\), we find:

\(x = \sqrt{5} - 1\)

(ii) To find the area of the shaded region, we need to integrate \(y = (2x - x^2)e^{\frac{1}{2}x}\) from \(x = 0\) to \(x = 2\).

Using integration by parts, we have:

\(\int (2x - x^2)e^{\frac{1}{2}x} \, dx\)

Integrating by parts twice, we obtain:

\((12x - 2x^2 - 24)e^{\frac{1}{2}x}\)

Evaluating from \(x = 0\) to \(x = 2\):

\([12(2) - 2(2)^2 - 24]e^{1} - [12(0) - 2(0)^2 - 24]e^{0}\)

\(= 24e - 8e - 24\)

\(= 24 - 8e\)

(a) Let \(u = \cos x\), then \(\frac{du}{dx} = -\sin x\) or \(dx = -\frac{du}{\sin x}\).

Using the double angle formula, \(\sin 2x = 2 \sin x \cos x = 2 \sin x \cdot u\).

Substitute into the integral:

\(\int \sin 2x \, e^{2 \cos x} \, dx = \int 2 \sin x \cdot u \, e^{2u} \cdot -\frac{du}{\sin x} = \int -2u e^{2u} \, du\).

Change the limits: when \(x = 0, u = \cos 0 = 1\) and when \(x = \pi, u = \cos \pi = -1\).

Thus, \(\int_{0}^{\pi} \sin 2x \, e^{2 \cos x} \, dx = \int_{1}^{-1} -2u e^{2u} \, du = \int_{-1}^{1} 2u e^{2u} \, du\).

(b) Integrate \(\int_{-1}^{1} 2u e^{2u} \, du\):

Let \(I = \int 2u e^{2u} \, du\).

Using integration by parts, let \(v = u\) and \(dw = 2e^{2u} \, du\), then \(dv = du\) and \(w = e^{2u}\).

\(I = u e^{2u} - \int e^{2u} \, du = u e^{2u} - \frac{1}{2} e^{2u} + C\).

Evaluate from \(-1\) to \(1\):

\(\left[ u e^{2u} - \frac{1}{2} e^{2u} \right]_{-1}^{1} = \left( 1 \cdot e^{2} - \frac{1}{2} e^{2} \right) - \left( -1 \cdot e^{-2} - \frac{1}{2} e^{-2} \right)\).

\(= \frac{1}{2} e + \frac{3}{2} e^{-2}\).

(a) Let \(u = \\sin x\), then \(du = \\cos x \, dx\). The integral becomes \(\int \\sin 2x \\cos^2 x \, dx = \int 2u(1-u^2) \, du\).

Integrate to get \(\int 2(u - u^3) \, du = 2 \left( \frac{u^2}{2} - \frac{u^4}{4} \right) = u^2 - \frac{u^4}{2}\).

Evaluate from \(u = 0\) to \(u = 1\):

\(\left[ u^2 - \frac{u^4}{2} \right]_0^1 = \left( 1 - \frac{1}{2} \right) - (0 - 0) = \frac{1}{2}\).

(b) Differentiate \(y = \\sin 2x \\cos^2 x\) using the product rule:

\(\frac{dy}{dx} = 2 \\cos 2x \\cos^2 x - 2 \\sin 2x \\cos x \\sin x\).

Set \(\frac{dy}{dx} = 0\) and solve:

\(2 \\cos 2x \\cos^2 x = 2 \\sin 2x \\cos x \\sin x\).

Using double angle formulas, simplify to \(4 \\sin^2 x = 1\), giving \(\\sin^2 x = \frac{1}{4}\).

Thus, \(\\sin x = \frac{1}{2}\), so \(x = \frac{1}{6}\pi\).