(a) To find the value of \(a\), we set \(y = 0\) in the equation \(y = \\sin \\sqrt{x}\). This gives \(\\sin \\sqrt{x} = 0\), which implies \(\\sqrt{x} = n\\pi\) for some integer \(n\). The smallest positive \(x\) occurs when \(\\sqrt{x} = \\pi\), so \(x = \\pi^2\). Thus, \(a = \\pi^2\).

(b) We use the substitution \(u = \\sqrt{x}\), which implies \(x = u^2\) and \(dx = 2u \, du\). The integral becomes:

\(\\int_0^{\\pi} \\sin u \, 2u \, du\)

Using integration by parts, let \(v = u\) and \(dw = \\sin u \, du\), then \(dv = du\) and \(w = -\\cos u\). The integration by parts formula \(\\int v \, dw = vw - \\int w \, dv\) gives:

\(-u \\cos u + \\int \\cos u \, du\)

The integral of \(\\cos u\) is \(\\sin u\), so the expression becomes:

\(-u \\cos u + \\sin u\)

Evaluating from \(u = 0\) to \(u = \\pi\):

\([-\\pi \\cos \\pi + \\sin \\pi] - [0 \\cos 0 + \\sin 0] = -\\pi(-1) + 0 = \\pi\)

Thus, the area is \(2\\pi\).

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

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

Set the derivative to zero to find critical points:

\(\cos x \cdot \cos 2x - \sin x \cdot 2\sin 2x = 0\).

Using the double angle formula \(\cos 2x = 2\cos^2 x - 1\), we simplify to:

\(6\sin^2 x = 1\) or \(6\cos^2 x = 5\) or \(5\tan^2 x = 1\).

Solving gives \(x = 0.421\) to 3 significant figures.

(b) Using the substitution \(u = \cos x\), we have \(du = -\sin x \, dx\).

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

Integrating, we get \(\frac{u}{3} - \frac{2}{3}u^3\).

Using limits \(u = 1\) and \(u = \frac{1}{\sqrt{2}}\), we evaluate:

\(\frac{1}{3}(\sqrt{2} - 1)\).

(a) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = \frac{x}{1 + 3x^4}\). We use the quotient rule to differentiate:

\(\frac{d}{dx} \left( \frac{x}{1 + 3x^4} \right) = \frac{(1 + 3x^4) \cdot 1 - x \cdot 12x^3}{(1 + 3x^4)^2} = \frac{1 + 3x^4 - 12x^4}{(1 + 3x^4)^2} = \frac{1 - 9x^4}{(1 + 3x^4)^2}\).

Set the derivative to zero to find critical points:

\(1 - 9x^4 = 0\)

\(9x^4 = 1\)

\(x^4 = \frac{1}{9}\)

\(x = \left( \frac{1}{9} \right)^{1/4} = \frac{1}{3^{1/2}} = \frac{1}{\sqrt{3}} \approx 0.577\).

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

Substitute \(x = \frac{u}{\sqrt{3}}\) into the integral:

\(\int_0^1 \frac{x}{1 + 3x^4} \, dx = \int_0^{\sqrt{3}} \frac{\frac{u}{\sqrt{3}}}{1 + u^2} \cdot \frac{du}{2\sqrt{3}x}\).

Since \(x = \frac{u}{\sqrt{3}}\), the integral becomes:

\(\int_0^{\sqrt{3}} \frac{1}{2\sqrt{3}(1 + u^2)} \, du\).

This is a standard integral of the form \(\frac{1}{a} \arctan(u)\), where \(a = 1\):

\(\frac{1}{2\sqrt{3}} \left[ \arctan(u) \right]_0^{\sqrt{3}} = \frac{1}{2\sqrt{3}} \left( \arctan(\sqrt{3}) - \arctan(0) \right)\).

\(\arctan(\sqrt{3}) = \frac{\pi}{3}\), so the area is:

\(\frac{1}{2\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\sqrt{3}\pi}{18}\).

(i) Let \(u = \\cos x\), then \(du = -\\sin x \, dx\). The integral becomes:

\(\int \\sin^3 x \\sqrt{\\cos x} \, dx = \int -u^{1/2} (1-u^2) \, du\).

Integrate to get:

\(-\left( \frac{2}{3} u^{3/2} + \frac{2}{7} u^{7/2} \right)\).

Change limits: when \(x = 0, u = 1\) and when \(x = \frac{1}{2} \pi, u = 0\).

Substitute limits:

\(-\left( \frac{2}{3} (0)^{3/2} + \frac{2}{7} (0)^{7/2} \right) + \left( \frac{2}{3} (1)^{3/2} + \frac{2}{7} (1)^{7/2} \right) = \frac{8}{21}\).

(ii) Differentiate \(y = \\sin^3 x \\sqrt{\\cos x}\) using the product and chain rules:

\(\frac{dy}{dx} = 3 \\sin^2 x \\cos x \\sqrt{\\cos x} - \frac{1}{2} \\sin^3 x \\frac{\\sin x}{\\sqrt{\\cos x}}\).

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

\(3 \\sin^2 x \\cos^2 x = \frac{1}{2} \\sin^4 x\).

Rearrange to find \(\tan^2 x = 6\).

Thus, \(x = \arctan(\\sqrt{6}) \approx 1.183\).

(i) To find the \(x\)-coordinate of \(M\), we first find the derivative of \(y = 5 \sin^2 x \cos^3 x\) using the product rule. Let \(u = \sin^2 x\) and \(v = \cos^3 x\). Then \(\frac{du}{dx} = 2 \sin x \cos x\) and \(\frac{dv}{dx} = -3 \cos^2 x \sin x\).

The derivative \(\frac{dy}{dx} = 5 \left( u \frac{dv}{dx} + v \frac{du}{dx} \right) = 5 \left( \sin^2 x (-3 \cos^2 x \sin x) + \cos^3 x (2 \sin x \cos x) \right)\).

Simplifying, \(\frac{dy}{dx} = 5 \left( -3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x \right)\).

Setting \(\frac{dy}{dx} = 0\), we get \(-3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x = 0\).

Factoring out \(\sin x \cos^2 x\), we have \(\sin x \cos^2 x (-3 \sin^2 x + 2 \cos^2 x) = 0\).

Thus, \(-3 \sin^2 x + 2 \cos^2 x = 0\) leads to \(3 \tan^2 x = 2\).

Solving for \(x\), \(\tan^2 x = \frac{2}{3}\), so \(x = \arctan \left( \sqrt{\frac{2}{3}} \right) \approx 0.685\).

(ii) Using the substitution \(u = \sin x\), \(du = \cos x \, dx\). The limits change from \(x = 0\) to \(x = \frac{1}{2} \pi\) to \(u = 0\) to \(u = 1\).

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

This simplifies to \(5 \int (u^2 - u^4) \, du\).

Integrating, \(5 \left( \frac{1}{3} u^3 - \frac{1}{5} u^5 \right)\) from 0 to 1.

Evaluating, \(5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5}{15} - \frac{3}{15} \right) = 5 \times \frac{2}{15} = \frac{10}{15} = \frac{2}{3}\).