Referred to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively such that $|\mathbf{a}| = 5$ and $|\mathbf{b}| = 3$.

1. Relate the magnitude of the vector to its scalar product with itself.

$|2\mathbf{b} - \mathbf{a}|^2 = 7^2$
$(2\mathbf{b} - \mathbf{a}) \cdot (2\mathbf{b} - \mathbf{a}) = 49$

2. Expand the scalar product, substitute the known magnitudes, and solve for $\mathbf{a}\cdot\mathbf{b}$.

$4(\mathbf{b}\cdot\mathbf{b}) - 2(\mathbf{a}\cdot\mathbf{b}) - 2(\mathbf{b}\cdot\mathbf{a}) + (\mathbf{a}\cdot\mathbf{a}) = 49$
$4|\mathbf{b}|^2 - 4(\mathbf{a}\cdot\mathbf{b}) + |\mathbf{a}|^2 = 49$

Substituting $|\mathbf{a}| = 5$ and $|\mathbf{b}| = 3$:
$4(3)^2 - 4(\mathbf{a}\cdot\mathbf{b}) + (5)^2 = 49$
$36 - 4(\mathbf{a}\cdot\mathbf{b}) + 25 = 49$
$61 - 4(\mathbf{a}\cdot\mathbf{b}) = 49$
$4(\mathbf{a}\cdot\mathbf{b}) = 12$
$\mathbf{a}\cdot\mathbf{b} = 3 \text{ (shown)}$

1. Evaluate the length of projection of $\overrightarrow{OA}$ onto $\overrightarrow{OB}$ by substituting the value of $\mathbf{a} \cdot \mathbf{b}$ from part (i) into the projection formula.

$\text{Length of projection} = \frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{b}|} = \frac{3}{3} = 1$

1. Find the position vector of $P$ using the ratio theorem.

Using the ratio theorem, since the point $P$ divides $AB$ in the ratio $AP:PB = 1:k$,
$\overrightarrow{OP} = \frac{k\overrightarrow{OA} + 1\overrightarrow{OB}}{1+k} = \frac{k\mathbf{a} + \mathbf{b}}{k+1}$

2. Find the sine of the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$.

From part (i), we have $\mathbf{a} \cdot \mathbf{b} = 3$. Let $\theta$ be the angle between vectors $\mathbf{a}$ and $\mathbf{b}$.
$|\mathbf{a}||\mathbf{b}|\cos\theta = 3$
$(5)(3)\cos\theta = 3 \implies \cos\theta = \frac{1}{5}$
Since $\theta$ is the angle between two vectors ($0 \le \theta \le \pi$), $\sin\theta \ge 0$.
$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$

3. Express the area of triangle $OAP$ using the vector product and simplify.

The area of triangle $OAP$ is given by $\frac{1}{2} |\mathbf{a} \times \mathbf{p}|$.
$\text{Area} = \frac{1}{2} \left| \mathbf{a} \times \left( \frac{k\mathbf{a} + \mathbf{b}}{k+1} \right) \right|$
Using the distributive property of the vector product and the fact that $\mathbf{a} \times \mathbf{a} = \mathbf{0}$:
$\text{Area} = \frac{1}{2(k+1)} |k(\mathbf{a} \times \mathbf{a}) + (\mathbf{a} \times \mathbf{b})| = \frac{1}{2(k+1)} |\mathbf{a} \times \mathbf{b}|$

4. Evaluate the magnitude of the vector product and show the final area.

We evaluate $|\mathbf{a} \times \mathbf{b}|$:
$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = (5)(3)\left(\frac{2\sqrt{6}}{5}\right) = 6\sqrt{6}$
Substituting this back into the area equation:
$\text{Area of } OAP = \frac{1}{2(k+1)}(6\sqrt{6}) = \frac{3\sqrt{6}}{k+1} \text{ (shown)}$