With reference to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively.

1. Identify the geometrical shape using the parallelogram law of vector addition

Since $\vec{OC} = \mathbf{a} + \mathbf{b} = \vec{OA} + \vec{OB}$, by the parallelogram law of vector addition, the points form a parallelogram.

1. Express the displacement vector $\vec{AM}$ in terms of $\mathbf{b}$.

From part (a), the position vector of $C$ is $\mathbf{a} + \mathbf{b}$.
$\vec{AC} = \vec{OC} - \vec{OA} = (\mathbf{a} + \mathbf{b}) - \mathbf{a} = \mathbf{b}$
Since $M$ lies on $AC$ and $AM : MC = 1 : 2$, we have:
$\vec{AM} = \frac{1}{1+2} \vec{AC} = \frac{1}{3}\mathbf{b}$

2. Calculate the area of triangle $OAM$ using the vector product and deduce $\alpha$.

The area of $\triangle OAM$ using sides $OA$ and $AM$ is given by $\frac{1}{2}|\vec{OA} \times \vec{AM}|$.
$\text{Area} = \frac{1}{2} \left| \mathbf{a} \times \frac{1}{3}\mathbf{b} \right|$
Using the properties of the vector product to extract the scalar:
$\text{Area} = \frac{1}{6} |\mathbf{a} \times \mathbf{b}|$
Comparing this to the given expression $\alpha\,|\mathbf{a} \times \mathbf{b}|$, we find:
$\alpha = \frac{1}{6}$

1. Determine the constant $\alpha$ from part (b) to express the area of triangle $OAM$.

Since point $C$ has position vector $\mathbf{a} + \mathbf{b}$, $OACB$ forms a parallelogram and $\vec{AC} = \mathbf{b}$.
Given $AM : MC = 1 : 2$, we have $\vec{AM} = \frac{1}{3}\mathbf{b}$.
Area of $\triangle OAM = \frac{1}{2}|\vec{OA} \times \vec{AM}| = \frac{1}{2}\left|\mathbf{a} \times \frac{1}{3}\mathbf{b}\right| = \frac{1}{6}|\mathbf{a} \times \mathbf{b}|$
So, $\alpha = \frac{1}{6}$.

2. Find the value of $|\mathbf{a} \times \mathbf{b}|$ using the given shortest distance from $A$ to $OB$.

The shortest distance $d$ from point $A$ to the line passing through $O$ and $B$ is given by $d = \frac{|\vec{OA} \times \vec{OB}|}{|\vec{OB}|}$.
$\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{b}|} = \sqrt{3}$
Since we are given $|\mathbf{b}| = 5$,
$|\mathbf{a} \times \mathbf{b}| = 5\sqrt{3}$

3. Calculate the exact area of triangle $OAM$.

Substituting $|\mathbf{a} \times \mathbf{b}| = 5\sqrt{3}$ into the result from part (b):
Area of $\triangle OAM = \frac{1}{6}(5\sqrt{3}) = \frac{5\sqrt{3}}{6}$

1. Find the position vector of $M$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

From part (a), $C$ has position vector $\mathbf{a} + \mathbf{b}$, so $\vec{AC} = (\mathbf{a} + \mathbf{b}) - \mathbf{a} = \mathbf{b}$.
Since $M$ lies on $AC$ with $AM : MC = 1 : 2$, we have $\vec{AM} = \frac{1}{3}\vec{AC} = \frac{1}{3}\mathbf{b}$.
$\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{3}\mathbf{b}$

2. Find the position vector of $N$ using the given ratio.

Point $N$ lies on $OM$ with $ON : NM = 3 : 1$, which means $\vec{ON} = \frac{3}{4}\vec{OM}$.
$\vec{ON} = \frac{3}{4}\left(\mathbf{a} + \frac{1}{3}\mathbf{b}\right) = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

3. Express $\vec{AN}$ in terms of $\vec{AB}$ to show collinearity.

We find the displacement vector $\vec{AN}$:
$\vec{AN} = \vec{ON} - \vec{OA} = \left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right) - \mathbf{a} = -\frac{1}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} = \frac{1}{4}(\mathbf{b} - \mathbf{a})$
Since $\vec{AB} = \mathbf{b} - \mathbf{a}$, we have:
$\vec{AN} = \frac{1}{4}\vec{AB}$
Because $\vec{AN}$ is a scalar multiple of $\vec{AB}$, the vectors are parallel. Since they share the common point $A$, the points $A$, $N$, and $B$ are collinear.

4. Deduce the ratio $AN : NB$.

Since $\vec{AN} = \frac{1}{4}\vec{AB}$, the length of $AN$ is $\frac{1}{4}$ of the total length of $AB$.
The remaining segment $NB$ takes up $\frac{3}{4}$ of the length.
Therefore, $AN : NB = \frac{1}{4} : \frac{3}{4} = 1 : 3$.