Relative to the origin $O$, the position vectors of points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively, where $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel vectors. The point $P$ on $OA$ is such that $OP : PA = 2 : 3$. The point $Q$ is such that $OPQB$ is a parallelogram.

1. Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$.

Given that $P$ is on $OA$ and $OP : PA = 2 : 3$, the total number of parts along $OA$ is $2 + 3 = 5$.

$\overrightarrow{OP} = \frac{2}{5}\overrightarrow{OA}$
$\overrightarrow{OP} = \frac{2}{5}\mathbf{a}$

2. Use the properties of a parallelogram to find $\overrightarrow{OQ}$.

Since $OPQB$ is a parallelogram, the adjacent sides are $\overrightarrow{OP}$ and $\overrightarrow{OB}$. By the parallelogram law of vector addition, the diagonal $\overrightarrow{OQ}$ from the origin is the sum of the adjacent side vectors.

$\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{OB}$
$\overrightarrow{OQ} = \frac{2}{5}\mathbf{a} + \mathbf{b}$

1. State the vector $\overrightarrow{OQ}$ derived from part (i).

Since $P$ is on $OA$ with $OP:PA = 2:3$, we have $\overrightarrow{OP} = \frac{2}{5}\mathbf{a}$.
As $OPQB$ is a parallelogram, $\overrightarrow{BQ} = \overrightarrow{OP} = \frac{2}{5}\mathbf{a}$.
Thus, $\overrightarrow{OQ} = \overrightarrow{OB} + \overrightarrow{BQ} = \mathbf{b} + \frac{2}{5}\mathbf{a}$.

2. Set up the expression for the area of triangle $OAQ$ using the cross product.

The area of a triangle defined by vectors $\overrightarrow{OA}$ and $\overrightarrow{OQ}$ is given by $\frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OQ}|$.
$\text{Area of } \triangle OAQ = \frac{1}{2} \left| \mathbf{a} \times \left( \frac{2}{5}\mathbf{a} + \mathbf{b} \right) \right|$

3. Expand the cross product to simplify the expression and find $k$.

Using the distributive property of the cross product:
$\text{Area} = \frac{1}{2} \left| \frac{2}{5}(\mathbf{a} \times \mathbf{a}) + (\mathbf{a} \times \mathbf{b}) \right|$
Since the cross product of any vector with itself is the zero vector ($\mathbf{a} \times \mathbf{a} = \mathbf{0}$):
$\text{Area} = \frac{1}{2} |\mathbf{0} + \mathbf{a} \times \mathbf{b}| = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|$
Comparing this to $k|\mathbf{a} \times \mathbf{b}|$, we find $k = \frac{1}{2}$.

1. Determine the ratio using the shared height of the triangles

Triangles $OPB$ and $OAB$ share the same vertex $B$ and their bases $OP$ and $OA$ lie on the same line. Therefore, the ratio of their areas is equal to the ratio of their bases.

Since $OP : PA = 2 : 3$, it follows that $OP : OA = 2 : (2 + 3) = 2 : 5$.

$\text{Area of } \triangle OPB : \text{Area of } \triangle OAB = 2 : 5$

1. State the formula relating the magnitude of a vector product to the magnitudes of the vectors and the angle between them.

$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta$

2. Substitute the given values into the formula.

Since $\mathbf{a} \times \mathbf{b}$ is a unit vector, $|\mathbf{a} \times \mathbf{b}| = 1$.
Substituting $|\mathbf{a}| = 2$ and $\theta = 60^\circ$:
$1 = (2)|\mathbf{b}|\sin 60^\circ$

3. Evaluate the exact trigonometric ratio and solve for $|\mathbf{b}|$.

$1 = 2|\mathbf{b}|\left(\frac{\sqrt{3}}{2}\right)$
$1 = \sqrt{3}|\mathbf{b}|$
$|\mathbf{b}| = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$