(i) Let \(u = \\cos x\), then \(du = -\\sin x \, dx\). The integral becomes \(\int \\sin x \\cos^2 2x \, dx\). Using the double angle formula, express \(\\cos 2x = 2u^2 - 1\). The integrand becomes \((2u^2 - 1)^2\). Change limits: when \(x = 0, u = 1\) and when \(x = \frac{1}{4}\pi, u = \frac{1}{\sqrt{2}}\). The integral is \(\int_{1/\sqrt{2}}^1 (2u^2 - 1)^2 \, du\). Substitute limits in an integral of the form \(au^5 + bu^3 + cu\) to obtain \(\frac{1}{15}(7 - 4\sqrt{2})\).

(ii) Differentiate \(y = \\sin x \\cos^2 2x\) using the product rule and chain rule. Set the derivative to zero and solve for \(x\). Use trigonometric identities to obtain \(10\\cos^2 x = 9\) or \(10\\sin^2 x = 1\). Solve to find \(x = 0.32\).

(i) Substitute \(u = \sin x\) and \(du = \cos x \, dx\). The integrand becomes \(e^{2u}\).

The indefinite integral is \(\frac{1}{2} e^{2u}\).

Using limits \(u = 0\) and \(u = 1\), the definite integral is \(\frac{1}{2}(e^2 - 1)\).

(ii) Use the chain rule or product rule to find the derivative: \(2\cos x \, e^{2\sin x} \cos x - e^{2\sin x} \sin x\).

Equate the derivative to zero and solve the quadratic equation in \(\sin x\).

Solve the 3-term quadratic to find \(x = 0.896\).

(i) To find the equation of the tangent, we first need the derivative of \(y = x^{\frac{1}{2}} \ln x\). Using the product rule, \(\frac{d}{dx}(x^{\frac{1}{2}} \ln x) = \ln x \cdot \frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}} \cdot \frac{1}{x} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}\).

At \(x = 1\), the derivative becomes \(\frac{\ln 1}{2} + 1 = 1\).

The point on the curve at \(x = 1\) is \((1, 0)\) since \(y = 1^{\frac{1}{2}} \ln 1 = 0\).

The equation of the tangent is \(y - 0 = 1(x - 1)\), which simplifies to \(y = x - 1\).

(ii) The volume of the solid obtained by rotating the region \(R\) about the x-axis is given by \(\pi \int_1^e (x^{\frac{1}{2}} \ln x)^2 \, dx\).

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

Integrating by parts, we have \(\frac{1}{2} x^2 (\ln x)^2 - \int x \ln x \, dx\).

For the second integral, use integration by parts again with \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

This gives \(\frac{1}{2} x^2 \ln x - \frac{1}{4} x^2\).

Substituting back, the integral becomes \(\frac{1}{2} x^2 (\ln x)^2 - \frac{1}{2} x^2 \ln x + \frac{1}{4} x^2\).

Evaluating from \(x = 1\) to \(x = e\), we get \(\frac{1}{4} \pi (e^2 - 1)\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = e^{-\frac{1}{2}x} \sqrt{1 + 2x}\). Differentiate \(y\) with respect to \(x\) using the product and chain rules. Set the derivative equal to zero and solve for \(x\). The derivative is:

\(\frac{dy}{dx} = e^{-\frac{1}{2}x} \left( \frac{1}{\sqrt{1 + 2x}} - \frac{x}{\sqrt{1 + 2x}} \right)\).

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

\(\frac{1}{\sqrt{1 + 2x}} - \frac{x}{\sqrt{1 + 2x}} = 0\).

This simplifies to \(1 = x\), so \(x = \frac{1}{2}\).

(ii) To find the volume of the solid obtained when \(R\) is rotated about the \(x\)-axis, use the formula for the volume of revolution:

\(V = \pi \int e^{-x}(1 + 2x) \, dx\).

Integrate by parts:

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

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

Evaluate from \(x = -\frac{1}{2}\) to \(x = 0\):

\(V = \pi \left[ -e^{0}(1 + 0) - 2e^{0} - (-e^{-\frac{1}{2}}(1 + 2(-\frac{1}{2})) - 2e^{-\frac{1}{2}}) \right]\).

\(= \pi (2\sqrt{e} - 3)\).

(a) To find the quotient and remainder when \(8x^3 + 4x^2 + 2x + 7\) is divided by \(4x^2 + 1\), perform polynomial division:

1. Divide the leading term \(8x^3\) by \(4x^2\) to get \(2x\).

2. Multiply \(2x\) by \(4x^2 + 1\) to get \(8x^3 + 2x\).

3. Subtract \(8x^3 + 2x\) from \(8x^3 + 4x^2 + 2x + 7\) to get \(4x^2 + 7\).

4. Divide \(4x^2\) by \(4x^2\) to get \(1\).

5. Multiply \(1\) by \(4x^2 + 1\) to get \(4x^2 + 1\).

6. Subtract \(4x^2 + 1\) from \(4x^2 + 7\) to get \(6\).

The quotient is \(2x + 1\) and the remainder is \(6\).

(b) The integral becomes:

\(\int_0^{\frac{1}{2}} \left(2x + 1 + \frac{6}{4x^2 + 1}\right) \, dx\)

Split the integral:

\(\int_0^{\frac{1}{2}} (2x + 1) \, dx + \int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx\)

1. \(\int_0^{\frac{1}{2}} (2x + 1) \, dx = \left[x^2 + x\right]_0^{\frac{1}{2}} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\)

2. \(\int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx = 3 \arctan(2x)\) evaluated from 0 to \(\frac{1}{2}\).

\(= 3 \left(\arctan(1) - \arctan(0)\right) = 3 \left(\frac{\pi}{4}\right) = \frac{3\pi}{4}\)

Combine results: \(\frac{3}{4} + \frac{3\pi}{4} = \frac{3}{4}(1 + \pi)\).