\( \mathbf{a)} \)
\( (i) \)
\( \overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2} \overrightarrow{AC} = \overrightarrow{OA} + \frac{1}{2} (\overrightarrow{OC} - \overrightarrow{OA}) \)
\( = a + \frac{1}{2} (c - a) = \frac{1}{2} a + \frac{1}{2} c \)

\( (ii) \)
\( b = \frac{5}{2} \overrightarrow{OM} = \frac{5}{2} \left( \frac{1}{2} a + \frac{1}{2} c \right) = \frac{5}{4} a + \frac{5}{4} c \)

\( \mathbf{b)} \)
\( (i) \)
\( |-10i + 24j| = \sqrt{(-10)^2 + 24^2} = \sqrt{676} = 26 \)
\( p = \frac{39}{26} (-10i + 24j) = \frac{3}{2} (-10i + 24j) = -15i + 36j \)

\( (ii) \)
\( 2p + q \) is parallel to the y-axis, so \( i \) component is zero.
Since \( 2p + q \) has a magnitude of 12, then \( 2p + q = 0i + 12j \)
\( 2(-15i + 36j) + xi + yj = 0i + 12j \)
\( -30i + x i = 0 \) and \( 72j + y j = 12 \)
\( x = 30 \) and \( y = -60 \)
\( q = 30i - 60j \)

\( (iii) \)
\( |q| = \sqrt{30^2 + (-60)^2} = \sqrt{30^2 + 4 \cdot 30^2} \)
\( = \sqrt{30^2 (1+4)} = 30 \sqrt{5} \)

(a) Given that \( \mathbf{p} = 2 \mathbf{i} - 5 \mathbf{j} \) and \( \mathbf{q} = \mathbf{i} - 3 \mathbf{j} \), we find \( 3\mathbf{p} - 4\mathbf{q} \):
\[ 3(2 \mathbf{i} - 5 \mathbf{j}) - 4(\mathbf{i} - 3 \mathbf{j}) \] Expanding: \[ (6 \mathbf{i} - 15 \mathbf{j}) - (4 \mathbf{i} - 12 \mathbf{j}) \] \[ (6 \mathbf{i} - 4 \mathbf{i}) + (-15 \mathbf{j} + 12 \mathbf{j}) \] \[ 2 \mathbf{i} - 3 \mathbf{j} \] Now, we find the magnitude:
\[ |2 \mathbf{i} - 3 \mathbf{j}| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}. \]
The unit vector is:
\[ \frac{2 \mathbf{i} - 3 \mathbf{j}}{\sqrt{13}}.\]

\( \mathbf{(i)} \)
\( b - c = \begin{pmatrix} 1 \\ 3 \end{pmatrix} - \begin{pmatrix} -5 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} \)
Since \( |a| = |b - c| \)
\( 2^2 + y^2 = 6^2 + (-2)^2 \)
\( y^2 = 6^2 \)
\( y = \pm 6 \)

\( \mathbf{(ii)} \)
\( 2b - c = \begin{pmatrix} 2 \\ 6 \end{pmatrix} - \begin{pmatrix} -5 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} \)
Then \( \lambda (2b - c) = \begin{pmatrix} 7 \lambda \\ \lambda \end{pmatrix} \)
\( \mu (b + c) + 4 (b - c) = \mu \begin{pmatrix} 1 - 5 \\ 3 + 5 \end{pmatrix} + 4 \begin{pmatrix} 6 \\ -2 \end{pmatrix} \)
\( = \mu \begin{pmatrix} -4 \\ 8 \end{pmatrix} + \begin{pmatrix} 24 \\ -8 \end{pmatrix} = \begin{pmatrix} -4 \mu + 24 \\ 8 \mu - 8 \end{pmatrix} \)
Since \( \mu (b+c) + 4 (b-c) = \lambda (2b-c) \)
\( -4 \mu + 24 = 7 \lambda \) and \( 8 \mu - 8 = \lambda \)
\( \begin{cases} 7 \lambda + 4 \mu = 24 \\ \lambda = 8 \mu - 8 \end{cases} \)
\( 56 \mu - 56 + 4 \mu = 24 \)
\( 60 \mu = 80 \)
\( \mu = \frac{4}{3} \)
\( \lambda = 8 \times \frac{4}{3} - 8 = \frac{8}{3} \)

(a) The given direction of vector \( \mathbf{v} \) is \( \mathbf{i} - 2 \mathbf{j} \).
The magnitude of \( \mathbf{i} - 2 \mathbf{j} \) is:
\( \sqrt{1^2 + (-2)^2} = \sqrt{5} \).
The unit vector in this direction is:
\( \frac{1}{\sqrt{5}} (\mathbf{i} - 2 \mathbf{j}) \).
Given that \( |\mathbf{v}| = 3\sqrt{5} \), we find \( \mathbf{v} \):
\( \mathbf{v} = 3\sqrt{5} \times \frac{1}{\sqrt{5}} (\mathbf{i} - 2 \mathbf{j}) = 3 \mathbf{i} - 6 \mathbf{j} \).

(b) The velocity vector \( \mathbf{w} \) makes an angle of \( 30^\circ \) with the positive \( x \)-axis and has magnitude \( 2 \).
Using trigonometric components:
\( \mathbf{w} = |\mathbf{w}| (\cos 30^\circ \mathbf{i} + \sin 30^\circ \mathbf{j}) \).
\( = 2 \left( \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \right) \).
\( = \sqrt{3} \mathbf{i} + 1 \mathbf{j} \).

(a) The vector \( \overrightarrow{A B} \) is given by:
\( \overrightarrow{A B} = \overrightarrow{O B} - \overrightarrow{O A} \).
\( = \mathbf{b} - \mathbf{a} \).

(b) The point \( P \) lies on \( O A \) such that \( O P = \frac{3}{4} O A \), so:
\( \overrightarrow{O P} = \frac{3}{4} \mathbf{a} \).
The point \( Q \) is the mid-point of \( A B \), so:
\( \overrightarrow{O Q} = \frac{1}{2} (\overrightarrow{O A} + \overrightarrow{O B}) = \frac{1}{2} (\mathbf{a} + \mathbf{b}) \).
The vector \( \overrightarrow{P Q} \) is:
\( \overrightarrow{P Q} = \overrightarrow{O Q} - \overrightarrow{O P} \).
\( = \frac{1}{2} (\mathbf{a} + \mathbf{b}) - \frac{3}{4} \mathbf{a} \).
\( = \frac{1}{2} \mathbf{a} + \frac{1}{2} \mathbf{b} - \frac{3}{4} \mathbf{a} \).
\( = \left( \frac{1}{2} - \frac{3}{4} \right) \mathbf{a} + \frac{1}{2} \mathbf{b} \).
\( = -\frac{1}{4} \mathbf{a} + \frac{1}{2} \mathbf{b} \).
(c) Given that \( n \overrightarrow{P Q} = \overrightarrow{Q R} \), we substitute \( \overrightarrow{P Q} \):
\( \overrightarrow{Q R} = n \left(-\frac{1}{4} \mathbf{a} + \frac{1}{2} \mathbf{b} \right) \).
\( = -\frac{n}{4} \mathbf{a} + \frac{n}{2} \mathbf{b} \).
(d) Given that \( \overrightarrow{B R} = k \mathbf{b} \), and using \( \overrightarrow{Q R} = \overrightarrow{B R} + \overrightarrow{Q B} \):
Since \( Q \) is the midpoint of \( A B \), we have \( \overrightarrow{Q B} = \frac{1}{2} \overrightarrow{A B} = \frac{1}{2} (\mathbf{b} - \mathbf{a}) \).
\( \overrightarrow{Q R} = k \mathbf{b} + \frac{1}{2} (\mathbf{b} - \mathbf{a}) \).
\( = k \mathbf{b} + \frac{1}{2} \mathbf{b} - \frac{1}{2} \mathbf{a} \).
\( = -\frac{1}{2} \mathbf{a} + \left( k + \frac{1}{2} \right) \mathbf{b} \).
(e) \( \frac{1}{2} (\mathbf{b} - \mathbf{a}) + k \mathbf{b} = n \left( \frac{1}{2} \mathbf{b} - \frac{1}{4} \mathbf{a} \right) \)
\( -\frac{1}{2} = -\frac{n}{4} \)
\( \frac{1}{2} + k = \frac{n}{2} \)
\( n = 2 \)
\( k = \frac{1}{2} \)