To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)

Given \(f(x) = \frac{3}{2x+5}\), we have:

\([f(x)]^2 = \left(\frac{3}{2x+5}\right)^2 = \frac{9}{(2x+5)^2}\)

The limits of integration are from \(x = 0\) to \(x = 2\).

Thus, the volume \(V\) is:

\(V = \pi \int_{0}^{2} \frac{9}{(2x+5)^2} \, dx\)

Integrating, we have:

\(V = \pi \left[ -\frac{9}{2(2x+5)} \right]_{0}^{2}\)

Evaluating the definite integral:

\(V = \pi \left( -\frac{9}{2(2 \times 2 + 5)} + \frac{9}{2(2 \times 0 + 5)} \right)\)

\(V = \pi \left( -\frac{9}{18} + \frac{9}{10} \right)\)

\(V = \pi \left( -0.5 + 0.9 \right)\)

\(V = \pi \times 0.4\)

\(V = 0.4\pi\)

Thus, the volume is \(0.4\pi\) (or approximately 1.26).

(i) To find the gradient of the curve at \(x = 2\), we first differentiate \(y = \frac{6}{3x - 2}\).

Using the chain rule, \(\frac{dy}{dx} = -6(3x - 2)^{-2} \times 3\).

Substitute \(x = 2\) into the derivative:

\(\frac{dy}{dx} = -6(3 \times 2 - 2)^{-2} \times 3 = -6(4)^{-2} \times 3 = -\frac{9}{8}\).

(ii) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int y^2 \, dx\).

\(V = \pi \int_{1}^{2} \left( \frac{6}{3x - 2} \right)^2 \, dx\).

\(V = \pi \int_{1}^{2} \frac{36}{(3x - 2)^2} \, dx\).

The integral of \(\frac{36}{(3x - 2)^2}\) is \(\frac{-36}{3x - 2}\).

Evaluate from 1 to 2:

\(\left[ \frac{-36}{3x - 2} \right]_{1}^{2} = \left( \frac{-36}{4} - \frac{-36}{1} \right) = -9 + 36 = 27\).

Thus, the volume is \(9\pi\).

(i) To find the area of the shaded region, integrate with respect to \(y\):

\(A = \int_1^2 \left( y^2 - 1 \right) \frac{1}{3} \, dy\)

\(= \left[ \frac{y^3}{9} - \frac{y}{3} \right]_1^2 = \frac{4}{9}\)

(ii) To find the volume when the region is rotated about the \(x\)-axis:

\(V = \pi \int_0^1 y^2 \, dx = \pi \int_0^1 (3x + 1) \, dx\)

\(= \pi \left[ \frac{3x^2}{2} + x \right]_0^1 = \pi \left( \frac{3}{2} + 1 \right) = \frac{5\pi}{2}\)

Subtract the volume of the cylinder: \(\pi \times 2^2 \times 1 = 4\pi\)

\(\text{Volume} = 4\pi - \frac{5\pi}{2} = 1.5\pi\)

(iii) To find the angle between the tangents, calculate the derivatives:

\(\frac{dy}{dx} = \frac{1}{2}(3x + 1)^{-\frac{1}{2}} \times 3\)

At \(x = 0\), \(m = \frac{3}{2}\). At \(x = 1\), \(m = \frac{3}{4}\).

Calculate the angles: \(\arctan(\frac{3}{2}) \approx 56.3^\circ\) and \(\arctan(\frac{3}{4}) \approx 36.9^\circ\).

The angle between the tangents is \(56.3^\circ - 36.9^\circ = 19.4^\circ\).

The volume \(V\) of the solid of revolution about the \(x\)-axis is given by the integral:

\(V = \pi \int_{1}^{4} y^2 \, dx\)

Substitute \(y = 3x^{\frac{1}{4}}\) into the integral:

\(V = \pi \int_{1}^{4} (3x^{\frac{1}{4}})^2 \, dx = \pi \int_{1}^{4} 9x^{\frac{1}{2}} \, dx\)

Integrate \(9x^{\frac{1}{2}}\):

\(\int 9x^{\frac{1}{2}} \, dx = 9 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = 6x^{\frac{3}{2}}\)

Evaluate from 1 to 4:

\(V = \pi \left[ 6x^{\frac{3}{2}} \right]_{1}^{4} = \pi \left( 6 \cdot 4^{\frac{3}{2}} - 6 \cdot 1^{\frac{3}{2}} \right)\)

Calculate \(4^{\frac{3}{2}} = 8\):

\(V = \pi (6 \cdot 8 - 6 \cdot 1) = \pi (48 - 6) = 42\pi\)

The volume \(V\) of the solid of revolution is given by:

\(V = \pi \int_0^1 \left( \frac{6}{5 - 2x} \right)^2 \, dx\)

First, simplify the integrand:

\(\left( \frac{6}{5 - 2x} \right)^2 = \frac{36}{(5 - 2x)^2}\)

Thus, the integral becomes:

\(V = \pi \int_0^1 \frac{36}{(5 - 2x)^2} \, dx\)

Let \(u = 5 - 2x\), then \(du = -2 \, dx\) or \(dx = -\frac{1}{2} \, du\).

Change the limits of integration: when \(x = 0\), \(u = 5\); when \(x = 1\), \(u = 3\).

Substitute and integrate:

\(V = \pi \int_5^3 \frac{36}{u^2} \left(-\frac{1}{2}\right) \, du\)

\(= -18\pi \int_5^3 \frac{1}{u^2} \, du\)

\(= -18\pi \left[ -\frac{1}{u} \right]_5^3\)

\(= 18\pi \left( \frac{1}{3} - \frac{1}{5} \right)\)

\(= 18\pi \left( \frac{5 - 3}{15} \right)\)

\(= 18\pi \left( \frac{2}{15} \right)\)

\(= \frac{36\pi}{15}\)

\(= \frac{12\pi}{5}\)

Thus, the volume obtained is \(\frac{12}{5} \pi\).