(a) To find the coordinates of \(M\), we need to find the maximum point of the function \(y = xe^{-\frac{1}{4}x^2}\). Differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = e^{-\frac{1}{4}x^2} - \frac{x^2}{2}e^{-\frac{1}{4}x^2}\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(e^{-\frac{1}{4}x^2} (1 - \frac{x^2}{2}) = 0\).

Since \(e^{-\frac{1}{4}x^2} \neq 0\), solve \(1 - \frac{x^2}{2} = 0\):

\(x^2 = 2\) \(\Rightarrow x = \sqrt{2}\).

Substitute \(x = \sqrt{2}\) back into the original equation to find \(y\):

\(y = \sqrt{2}e^{-\frac{1}{4}(\sqrt{2})^2} = \sqrt{2}e^{-\frac{1}{2}}\).

Thus, the coordinates of \(M\) are \(\left( \sqrt{2}, \sqrt{2}e^{-\frac{1}{2}} \right)\).

(b) To find the area under the curve from \(x = 0\) to \(x = 3\), use the substitution \(x = \sqrt{u}\), so \(dx = \frac{1}{2\sqrt{u}} du\).

The integral becomes:

\(\int xe^{-\frac{1}{4}x^2} \, dx = \int \frac{1}{2} e^{-\frac{1}{4}u} \, du\).

Change the limits: when \(x = 0, u = 0\) and when \(x = 3, u = 9\).

Integrate:

\(\frac{1}{2} \int e^{-\frac{1}{4}u} \, du = -2e^{-\frac{1}{4}u} \bigg|_0^9\).

Calculate the definite integral:

\(-2e^{-\frac{9}{4}} + 2e^0 = 2 - 2e^{-\frac{9}{4}}\).

Thus, the exact area of the shaded region is \(2 - 2e^{-\frac{9}{4}}\).

(i) To find the \(x\)-coordinate of \(M\), we first differentiate \(y = \\sin^2 2x \\cos x\) using the product rule:

\(\frac{dy}{dx} = 4 \\sin 2x \\cos 2x \\cos x - \\sin^2 2x \\sin x\).

Set the derivative to zero and use a double angle formula to simplify:

\(10 \\cos^3 x = 6 \\cos x\).

Solve for \(x\) to obtain \(x = 0.685\).

(ii) Using the substitution \(u = \\sin x\), express the integral in terms of \(u\) and \(du\):

\(du = \\pm \\cos x \, dx\).

The integral becomes \(\int 4u^2 (1-u^2) \, du\).

Use limits \(u = 0\) and \(u = 1\) to evaluate the integral:

\(\frac{8}{15}\) (or 0.533).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of the function \(y = 5 \sin^3 x \cos^2 x\). First, find the derivative using the product and chain rule:

\(\frac{dy}{dx} = 15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x\).

Set the derivative to zero to find critical points:

\(15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x = 0\).

Factor out common terms:

\(5 \sin^2 x \cos x (3 \cos^2 x - 2 \sin^2 x) = 0\).

Solving \(3 \cos^2 x - 2 \sin^2 x = 0\) gives \(3 \cos^2 x = 2 \sin^2 x\).

Using \(\sin^2 x + \cos^2 x = 1\), solve for \(x\):

\(2 \tan^2 x = 3\) leads to \(x = 0.886\) radians.

(ii) To find the area of the shaded region, use the substitution \(u = \cos x\), so \(du = -\sin x \, dx\).

The integral becomes:

\(\int 5 \sin^3 x \cos^2 x \, dx = \int 5 (1-u^2) u^2 (-du)\).

\(= \int 5 (u^2 - u^4) \, du\).

Integrate and use limits \(u = 1\) and \(u = 0\) (corresponding to \(x = 0\) and \(x = \frac{1}{2} \pi\)):

\(\int 5 (u^2 - u^4) \, du = 5 \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1\).

\(= 5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5 - 3}{15} \right) = \frac{2}{3}\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of \(y = x^2 \sqrt{1-x^2}\). Differentiate \(y\) with respect to \(x\) using the product and chain rule:

\(y = x^2 (1-x^2)^{1/2}\)

\(\frac{dy}{dx} = 2x \sqrt{1-x^2} + x^2 \cdot \frac{-x}{\sqrt{1-x^2}}\)

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

\(2x \sqrt{1-x^2} = \frac{x^3}{\sqrt{1-x^2}}\)

\(2x(1-x^2) = x^3\)

\(2x - 2x^3 = x^3\)

\(2x = 3x^3\)

\(x^2 = \frac{2}{3}\)

\(x = \sqrt{\frac{2}{3}}\)

(ii) Use the substitution \(x = \sin \theta\), then \(dx = \cos \theta \ d\theta\). The integral becomes:

\(A = \int_0^1 x^2 \sqrt{1-x^2} \ dx = \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \ d\theta\)

\(= \int_0^{\frac{\pi}{2}} \frac{1}{4} \sin^2 2\theta \ d\theta\)

\(= \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2 2\theta \ d\theta\)

(iii) To find the exact value of \(A\), integrate:

\(\int \sin^2 2\theta \ d\theta = \int \frac{1 - \cos 4\theta}{2} \ d\theta\)

\(= \frac{1}{2} \left( \theta - \frac{1}{4} \sin 4\theta \right)\)

Evaluate from \(0\) to \(\frac{\pi}{2}\):

\(A = \frac{1}{4} \left[ \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) \right] = \frac{1}{16} \pi\)

(a) To find the maximum point \(M\), we first differentiate \(y = (x + 5) \sqrt{3 - 2x}\) using the product rule:

\(\frac{d}{dx}[(x+5)(3-2x)^{1/2}] = (x+5) \cdot \frac{1}{2}(3-2x)^{-1/2}(-2) + \sqrt{3-2x}\).

Set the derivative to zero to find critical points:

\(-(x+5)(3-2x)^{-1/2} + \sqrt{3-2x} = 0\).

Simplifying gives \((x+5) = (3-2x)\).

Solving \(x + 5 = 3 - 2x\) gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the original equation to find \(y\):

\(y = \left( \frac{2}{3} + 5 \right) \sqrt{3 - 2 \times \frac{2}{3}} = \frac{2 \sqrt{13}}{3 \sqrt{3}}\).

Thus, the coordinates of \(M\) are \(\left( \frac{2}{3}, \frac{2 \sqrt{13}}{3 \sqrt{3}} \right)\).

(b) Using the substitution \(u = 3 - 2x\), we have \(du = -2 dx\) or \(dx = -\frac{1}{2} du\).

The limits of integration change from \(x = -5\) to \(u = 13\) and \(x = \frac{3}{2}\) to \(u = 0\).

The integral becomes:

\(\int_{13}^{0} (3-u)^{1/2} \cdot \frac{1}{2} du\).

Integrate to find the area:

\(-\frac{1}{2} \int_{13}^{0} (3-u)^{1/2} du = -\frac{1}{2} \left[ \frac{2}{3}(3-u)^{3/2} \right]_{13}^{0}\).

Evaluate the definite integral:

\(-\frac{1}{3} \left[ (3-0)^{3/2} - (3-13)^{3/2} \right] = \frac{169 \sqrt{13}}{15}\).