(i) To find the equation of the tangent at \(A\), we first find the derivative of the curve:

\(\frac{dy}{dx} = \frac{-10}{(2x+1)^2} \times 2\).

At \(A\), \(y = 0\), so \(x = 2\).

The slope \(m\) at \(x = 2\) is \(\frac{-4}{5}\).

The equation of the tangent is \(y = \frac{-4}{5}(x - 2)\).

Rearranging gives \(5y + 4x = 8\).

(ii) The point \(C\) is \((0, 1.6)\).

The distance \(AC\) is \(\sqrt{(1.6)^2 + 2^2} = 2.56\).

(i) Differentiate the curve: \(\frac{dy}{dx} = 4 - 2x\).

At \(x = 3\), the gradient \(m = -2\).

The gradient of the normal is \(\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

Equation of the normal: \(y - 6 = \frac{1}{2}(x - 3)\).

Simplify to get \(2y = x + 9\).

(ii) The normal meets the axes at \((0, \frac{9}{2})\) and \((-9, 0)\).

Mid-point formula: \(\left(\frac{-9 + 0}{2}, \frac{0 + \frac{9}{2}}{2}\right) = \left(-\frac{9}{2}, \frac{9}{4}\right)\).

(iii) Set \(2y = x + 9\) equal to \(y = 4x - x^2 + 3\).

Substitute to get \(2(4x - x^2 + 3) = x + 9\).

Simplify to \(2x^2 - 7x + 3 = 0\).

Solving gives \(x = \frac{1}{2}\).

Substitute back to find \(y = \frac{43}{4}\).

Coordinates are \(\left(\frac{1}{2}, \frac{43}{4}\right)\).

(i) To find the equation of the normal at \(A\), first find the derivative of the curve \(y = 2 - \frac{18}{2x+3}\).

\(\frac{dy}{dx} = 18(2x+3)^{-2} \times 2\).

At \(x = 3\), the slope \(m\) of the tangent is \(\frac{4}{9}\).

The slope of the normal is \(-\frac{9}{4}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y = -\frac{9}{4}(x-3)\).

Rearranging gives \(4y + 9x = 27\).

(ii) The normal meets the y-axis at \((0, 6\frac{3}{4})\).

The curve meets the y-axis at \((0, -4)\).

Thus, the length of \(BC\) is \(10\frac{3}{4}\).

(i) Differentiate \(y = 5 - \frac{8}{x}\) to find the gradient of the tangent: \(\frac{dy}{dx} = \frac{8}{x^2}\).

At \(x = 2\), \(\frac{dy}{dx} = \frac{8}{4} = 2\). The gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y - 1 = -\frac{1}{2}(x - 2)\), which simplifies to \(2y + x = 4\).

(ii) Solve the simultaneous equations \(2y + x = 4\) and \(y = 5 - \frac{8}{x}\).

Substitute \(y = \frac{4 - x}{2}\) into \(y = 5 - \frac{8}{x}\) to get \(\frac{4 - x}{2} = 5 - \frac{8}{x}\).

Multiply through by \(2x\) to clear fractions: \((4 - x)x = 10x - 16\).

Simplify to \(x^2 + 6x - 16 = 0\). Solving gives \(x = -8\) or \(x = 2\). Since \(x = 2\) is point \(P\), \(Q\) is \((-8, 6)\).

(iii) Use the distance formula to find \(PQ\): \(\sqrt{(2 - (-8))^2 + (1 - 6)^2} = \sqrt{10^2 + 5^2} = \sqrt{125}\).

The length of \(PQ\) is approximately \(11.2\).

(i) Differentiate \(y = 2x + \frac{8}{x^2}\) with respect to \(x\):

\(\frac{dy}{dx} = 2 - \frac{16}{x^3}\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{48}{x^4}\).

(ii) Set \(\frac{dy}{dx} = 0\) to find the stationary point:

\(2 - \frac{16}{x^3} = 0 \Rightarrow x = 2\).

Substitute \(x = 2\) into the original equation:

\(y = 2(2) + \frac{8}{2^2} = 4 + 2 = 6\).

Stationary point is \((2, 6)\).

Since \(\frac{d^2y}{dx^2}\) is positive, the stationary point is a minimum.

(iii) At \(x = -2\), \(\frac{dy}{dx} = 4\), so the gradient of the tangent is 4.

The gradient of the normal is \(-\frac{1}{4}\).

Equation of the normal: \(y + 2 = -\frac{1}{4}(x + 2)\).

Set \(y = 0\) to find the x-intercept:

\(0 + 2 = -\frac{1}{4}(x + 2) \Rightarrow x = -10\).

The normal intersects the x-axis at \((-10, 0)\).