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

Using the product rule, the derivative is:

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

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

\(2x e^{2-x} - x^2 e^{2-x} = 0\).

Factor out \(e^{2-x}\):

\(e^{2-x} (2x - x^2) = 0\).

Since \(e^{2-x} \neq 0\), we solve \(2x - x^2 = 0\).

\(x(2 - x) = 0\).

Thus, \(x = 0\) or \(x = 2\). Since \(M\) is a maximum point, \(x = 2\).

(ii) To find \(\int_0^2 x^2 e^{2-x} \, dx\), use integration by parts twice.

First integration by parts:

Let \(u = x^2\), \(dv = e^{2-x} \, dx\).

Then \(du = 2x \, dx\), \(v = -e^{2-x}\).

\(\int x^2 e^{2-x} \, dx = -x^2 e^{2-x} + \int 2x e^{2-x} \, dx\).

Second integration by parts on \(\int 2x e^{2-x} \, dx\):

Let \(u = 2x\), \(dv = e^{2-x} \, dx\).

Then \(du = 2 \, dx\), \(v = -e^{2-x}\).

\(\int 2x e^{2-x} \, dx = -2x e^{2-x} + \int 2 e^{2-x} \, dx\).

\(\int 2 e^{2-x} \, dx = -2 e^{2-x}\).

Thus, \(\int x^2 e^{2-x} \, dx = -x^2 e^{2-x} - 2x e^{2-x} - 2 e^{2-x}\).

Evaluate from 0 to 2:

\(\left[ -x^2 e^{2-x} - 2x e^{2-x} - 2 e^{2-x} \right]_0^2\).

At \(x = 2\), the expression is \(-4e^0 - 4e^0 - 2e^0 = -10\).

At \(x = 0\), the expression is \(0\).

Therefore, the integral is \(2e^2 - 10\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = x\) and \(v = \cos \frac{1}{2}x\). Then \(u' = 1\) and \(v' = -\frac{1}{2} \sin \frac{1}{2}x\).

Thus, \(\frac{dy}{dx} = 1 \cdot \cos \frac{1}{2}x + x \cdot \left(-\frac{1}{2} \sin \frac{1}{2}x\right) = \cos \frac{1}{2}x - \frac{1}{2}x \sin \frac{1}{2}x\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) again using the product rule.

\(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{2} \left( \cos \frac{1}{2}x + \frac{1}{2}x \sin \frac{1}{2}x \right)\).

Simplify to get \(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{4}x \cos \frac{1}{2}x - \frac{1}{2} \sin \frac{1}{2}x\).

Substitute into \(4 \frac{d^2y}{dx^2} + y + 4 \sin \frac{1}{2}x = 0\) to verify.

(ii) To find the area, integrate \(y = x \cos \frac{1}{2}x\) from \(0\) to \(\pi\).

\(\int_0^\pi x \cos \frac{1}{2}x \, dx = \left[ 2x \sin \frac{1}{2}x - 4 \cos \frac{1}{2}x \right]_0^\pi\).

Evaluate: \(2\pi - 4\).

(i) To find the minimum point \(M\), we first find the derivative of \(y = x^2 \ln x\) using the product rule:

\(\frac{dy}{dx} = 2x \ln x + x\).

Set the derivative to zero to find critical points:

\(2x \ln x + x = 0\).

\(x(2 \ln x + 1) = 0\).

Since \(x = 0\) is not in the domain, solve \(2 \ln x + 1 = 0\):

\(\ln x = -\frac{1}{2}\).

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

Substitute back to find \(y\):

\(y = (e^{-\frac{1}{2}})^2 \ln(e^{-\frac{1}{2}}) = e^{-1}(-\frac{1}{2}) = -\frac{1}{2}e^{-1}\).

Thus, \(M \left( e^{-\frac{1}{2}}, -\frac{1}{2}e^{-1} \right)\).

(ii) To find the area of the shaded region, integrate \(y = x^2 \ln x\) from \(x = 1\) to \(x = e\):

Use integration by parts:

Let \(u = \ln x\), \(dv = x^2 dx\).

Then \(du = \frac{1}{x} dx\), \(v = \frac{x^3}{3}\).

\(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\).

\(= \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx\).

\(= \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).

Evaluate from \(x = 1\) to \(x = e\):

\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^e\).

\(= \left( \frac{e^3}{3} \cdot 1 - \frac{1}{9} e^3 \right) - \left( \frac{1}{3} \cdot 0 - \frac{1}{9} \right)\).

\(= \frac{e^3}{3} - \frac{1}{9} e^3 + \frac{1}{9}\).

\(= \frac{1}{9}(2e^3 + 1)\).

(i) To find the area under the curve \(y = x^2 e^{-x}\) from \(x = 0\) to \(x = 3\), we integrate:

\(\int_0^3 x^2 e^{-x} \, dx\)

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

\(\int x^2 e^{-x} \, dx = -x^2 e^{-x} + \int 2x e^{-x} \, dx\)

Apply integration by parts again to \(\int 2x e^{-x} \, dx\):

Let \(u = 2x\) and \(dv = e^{-x} \, dx\). Then \(du = 2 \, dx\) and \(v = -e^{-x}\).

\(\int 2x e^{-x} \, dx = -2x e^{-x} + \int 2 e^{-x} \, dx\)

\(= -2x e^{-x} - 2e^{-x}\)

Substitute back:

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

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \right]_0^3\)

At \(x = 3\):

\(-9e^{-3} - 6e^{-3} - 2e^{-3} = -17e^{-3}\)

At \(x = 0\):

\(0 - 0 - 2 = -2\)

Thus, the area is:

\(-(-2) - (-17e^{-3}) = 2 - \frac{17}{e^3}\)

(ii) To find the maximum point, differentiate \(y = x^2 e^{-x}\):

\(\frac{dy}{dx} = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2)\)

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

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

\(2x - x^2 = 0\)

\(x(2 - x) = 0\)

\(x = 0\) or \(x = 2\). Since \(x = 0\) is not a maximum, \(x = 2\) is the maximum point.

(iii) The tangent at \(P\) passes through the origin. The equation of the tangent at \(x = a\) is:

\(y - a^2 e^{-a} = (2a - a^2)e^{-a}(x - a)\)

Substitute \((0,0)\):

\(0 - a^2 e^{-a} = (2a - a^2)e^{-a}(0 - a)\)

\(a^2 = a^2(2 - a)\)

\(2 - a = 1\)

\(a = 1\). Thus, \(x = 1\).

(i) To find the coordinates of \(M\), we first find the derivative of \(y = x^3 \ln x\) using the product rule:

\(\frac{dy}{dx} = 3x^2 \ln x + x^2\).

Set the derivative to zero to find the critical points:

\(3x^2 \ln x + x^2 = 0\).

Factor out \(x^2\):

\(x^2 (3 \ln x + 1) = 0\).

Since \(x \neq 0\), solve \(3 \ln x + 1 = 0\):

\(\ln x = -\frac{1}{3}\).

\(x = e^{-1/3}\).

Substitute \(x = e^{-1/3}\) back into the original equation to find \(y\):

\(y = (e^{-1/3})^3 \ln(e^{-1/3}) = -\frac{1}{3e}\).

Thus, the coordinates of \(M\) are \(\left( e^{-1/3}, -\frac{1}{3e} \right)\).

(ii) To find the area of the shaded region, integrate \(y = x^3 \ln x\) from \(x = 1\) to \(x = 2\):

\(\int_1^2 x^3 \ln x \, dx\).

Using integration by parts, let \(u = \ln x\) and \(dv = x^3 \, dx\):

\(du = \frac{1}{x} \, dx\), \(v = \frac{x^4}{4}\).

\(\int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx\).

\(= \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx\).

\(= \frac{x^4}{4} \ln x - \frac{1}{16} x^4 + C\).

Evaluate from \(x = 1\) to \(x = 2\):

\(\left[ \frac{x^4}{4} \ln x - \frac{1}{16} x^4 \right]_1^2\).

\(= \left( \frac{16}{4} \ln 2 - \frac{1}{16} \cdot 16 \right) - \left( \frac{1}{4} \ln 1 - \frac{1}{16} \right)\).

\(= 4 \ln 2 - 1 - 0 + \frac{1}{16}\).

\(= 4 \ln 2 - \frac{15}{16}\).