First, find the derivative of the curve \(y = \frac{4}{\sqrt{x}}\). Rewrite it as \(y = 4x^{-0.5}\).

The derivative is \(\frac{dy}{dx} = -2x^{-1.5}\).

At the point \((4, 2)\), the gradient of the tangent is \(\frac{dy}{dx} = -\frac{1}{4}\).

The gradient of the normal is the negative reciprocal, so \(m = 4\).

The equation of the normal is \(y - 2 = 4(x - 4)\).

Simplify to get \(y = 4x - 14\).

Find where the normal meets the \(x\)-axis by setting \(y = 0\):

\(0 = 4x - 14\)

\(x = 3.5\)

So, \(P(3.5, 0)\).

Find where the normal meets the \(y\)-axis by setting \(x = 0\):

\(y = -14\)

So, \(Q(0, -14)\).

The length of \(PQ\) is given by:

\(\sqrt{(3.5 - 0)^2 + (0 + 14)^2} = \sqrt{3.5^2 + 14^2}\)

\(= \sqrt{12.25 + 196} = \sqrt{208.25}\)

\(\approx 14.4\)

(a) To find the coordinates of \(P\), equate the line and curve equations: \(mx + c = -\frac{m}{x}\). Rearrange to get \(mx^2 + cx + m = 0\). For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\), giving \(c^2 - 4m^2 = 0\). Solving gives \(c = \pm 2m\). Substitute \(c = 2m\) into the quadratic: \(mx^2 + 2mx + m = 0\), which simplifies to \((x+1)^2 = 0\), so \(x = -1\). Substituting back, \(y = m\), so \(P = (-1, m)\).

(b) The equation of the normal at \(P\) is \(y - m = -\frac{1}{m}(x + 1)\). Simplifying gives \(y = -\frac{1}{m}x + \frac{1}{m} + m\). Equate this to the curve equation \(y = -\frac{m}{x}\) to find \(x\): \(-\frac{1}{m}x + \frac{1}{m} + m = -\frac{m}{x}\). Rearrange to get \(x^2(1 - m^2) - m^2 = 0\). Solving gives \(x = m^2\). Substitute back to find \(y\): \(y = -\frac{1}{m}\), so \(Q = (m^2, -\frac{1}{m})\).

(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

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

\(\frac{2}{\sqrt{x}} = 1\)

\(\sqrt{x} = 2\)

\(x = 4\)

Substitute \(x = 4\) into the curve equation to find \(y\):

\(y = 6\)

Thus, the coordinates of the stationary point are \((4, 6)\).

(iii) Differentiate \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{3}{x^2}\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

(iv) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = -\frac{1}{3}\).

The gradient of the normal is \(m = 3\) (since \(m_1 m_2 = -1\)).

The angle \(\theta\) is given by \(\tan \theta = 3\), so \(k = 3\).

(i) To find the equation of the tangent, we first need the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{3x-4}\).

The derivative is \(\frac{dy}{dx} = -4(3x-4)^{-2} \times 3\).

Substituting \(x = 2\) into the derivative gives the gradient \(m = -3\).

The equation of the tangent at \(P(2, 2)\) is \(y - 2 = -3(x - 2)\).

(ii) The tangent makes an angle \(\theta\) with the \(x\)-axis, where \(\tan \theta = \pm (-3)\).

This gives \(\theta = \pm 108.4^\circ\) (or \(\pm 71.6^\circ\)).

(a) To find the \(x\)-coordinates of \(A\) and \(B\), solve the simultaneous equations:

\(2y = x + 5\)

\(y = x^2 - 4x + 7\)

Substitute \(y = \frac{x + 5}{2}\) into the quadratic equation:

\(\frac{x + 5}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + 5 = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 9 = 0\)

Factorize or use the quadratic formula to solve:

\(x = 3 \text{ or } x = \frac{1}{2}\)

(b) To find the equation of the tangent to the curve at \(B\), first find the derivative:

\(\frac{dy}{dx} = 2x - 4\)

At \(x = 3\), the gradient \(m\) is:

\(m = 2(3) - 4 = 2\)

The equation of the tangent at \(x = 3\) is:

\(y - 4 = 2(x - 3)\)

(c) To find the acute angle between the tangent and the line \(2y = x + 5\), first find the gradients:

The gradient of the line \(2y = x + 5\) is \(\frac{1}{2}\).

The angle \(\theta_1\) with the \(x\)-axis for the tangent is:

\(\arctan(2) \approx 63.4^\circ\)

The angle \(\theta_2\) with the \(x\)-axis for the line is:

\(\arctan\left(\frac{1}{2}\right) \approx 26.6^\circ\)

The acute angle between them is:

\(63.4^\circ - 26.6^\circ = 36.8^\circ\)

Rounded to 1 decimal place, the angle is \(37^\circ\).