First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = 2\sqrt{3x+4} - x\).

Using the chain rule, \(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\).

Substitute \(x = 4\) into \(\frac{dy}{dx}\) to find the gradient of the tangent at the point (4, 4):

\(\frac{dy}{dx} = \frac{1}{4}\).

The gradient of the normal is the negative reciprocal of the gradient of the tangent:

\(m = -\frac{1}{\frac{1}{4}} = 4\).

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), with \(m = 4\) and point (4, 4):

\(y - 4 = 4(x - 4)\).

Simplify to find the equation of the normal:

\(y = 4x - 16 + 4\).

Thus, \(y = 4x - 12\).

(i) To find the points of intersection, set \(4x^{\frac{1}{2}} = x + 3\). Rearrange to get \((x^{\frac{1}{2}})^2 - 4x^{\frac{1}{2}} + 3 = 0\) or \(x^2 - 10x + 9 = 0\). Solving gives \(x = 1\) or \(x = 9\). For \(x = 1\), \(y = 4\); for \(x = 9\), \(y = 12\). The points are \(A(1, 4)\) and \(B(9, 12)\). The length \(AB = \sqrt{(9-1)^2 + (12-4)^2} = \sqrt{128} = 8\sqrt{2} \approx 11.3\).

(ii) The gradient of \(AB\) is \(1\). The derivative of the curve is \(\frac{dy}{dx} = 2x^{-\frac{1}{2}}\). Set \(2x^{-\frac{1}{2}} = 1\) to find \(x = 4\). Then \(y = 4 \times 4^{\frac{1}{2}} = 8\). So, \(T(4, 8)\).

(iii) The equation of the normal at \(T\) is \(y - 8 = -1(x - 4)\) or \(y = -x + 12\). Solve \(-x + 12 = x + 3\) to find \(x = \frac{9}{2}\). Substitute into \(y = x + 3\) to get \(y = \frac{15}{2}\). The intersection point is \(\left( \frac{9}{2}, \frac{15}{2} \right)\).

(i) To find the equation of line \(AB\), calculate the gradient: \(\text{Gradient of } AB = \frac{2 - 0}{5 - 1} = \frac{1}{2}\).

The equation of the line is \(y = \frac{1}{2}x - \frac{1}{2}\).

(ii) The derivative of the curve \(y = (x - 1)^{\frac{1}{2}}\) is \(\frac{dy}{dx} = \frac{1}{2}(x - 1)^{-\frac{1}{2}}\).

Set \(\frac{dy}{dx} = \frac{1}{2}\) to find the tangent parallel to \(AB\):

\(\frac{1}{2}(x - 1)^{-\frac{1}{2}} = \frac{1}{2}\) gives \(x = 2\), \(y = 1\).

The equation of the tangent is \(y - 1 = \frac{1}{2}(x - 2)\), simplifying to \(y = \frac{1}{2}x\).

(iii) The perpendicular distance \(d\) between the line \(AB\) and the tangent is given by:

\(d = \sin(26.5^\circ) = 0.45\).

To find the equation of the tangent, we first need to find the derivative \(\frac{dy}{dx}\) of the curve.

The given equation is \(y = 2x^{\frac{3}{2}} - 3x - 4x^{\frac{1}{2}} + 4\).

Differentiate term by term:

\(\frac{dy}{dx} = \frac{d}{dx}(2x^{\frac{3}{2}}) - \frac{d}{dx}(3x) - \frac{d}{dx}(4x^{\frac{1}{2}}) + \frac{d}{dx}(4)\).

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

Now, evaluate \(\frac{dy}{dx}\) at \(x = 4\):

\(\frac{dy}{dx} \bigg|_{x=4} = 3(4)^{\frac{1}{2}} - 3 - 2(4)^{-\frac{1}{2}}\).

\(\frac{dy}{dx} \bigg|_{x=4} = 6 - 3 - 1 = 2\).

The gradient of the tangent at \(x = 4\) is 2.

The point of tangency is (4, 0). Using the point-slope form of the equation of a line:

\(y - y_1 = m(x - x_1)\), where \(m = 2\), \(x_1 = 4\), and \(y_1 = 0\).

\(y - 0 = 2(x - 4)\).

Thus, the equation of the tangent is \(y = 2(x - 4)\).

The gradient of the line 3y + x = 25 is \\(-\frac{1}{3}\\). Since this line is normal to the curve, the gradient of the tangent to the curve at the point of intersection is 3.

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

Set the gradient equal to 3: \\(2x - 5 = 3\\).

Solve for \\(x\\):

\\(2x - 5 = 3\\)

\\(2x = 8\\)

\\(x = 4\\)

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

\\(3y + 4 = 25\\)

\\(3y = 21\\)

\\(y = 7\\)

Substitute \\(x = 4\\) and \\(y = 7\\) into the curve equation to find \\(k\\):

\\(7 = 4^2 - 5(4) + k\\)

\\(7 = 16 - 20 + k\\)

\\(7 = -4 + k\\)

\\(k = 11\\)