The points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ with respect to origin $O$, where $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel.

1. Express the displacement vector $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$

2. Use the collinearity of $A$, $B$, and $C$ to establish a vector equation.

Since $B$ lies on the line segment $AC$, the vectors $\overrightarrow{BC}$ and $\overrightarrow{AB}$ are parallel, so $\overrightarrow{BC} = k\overrightarrow{AB}$ for some scalar $k$.
$5\mathbf{b} - \mu\mathbf{a} = k(\mathbf{b} - \mathbf{a})$
$5\mathbf{b} - \mu\mathbf{a} = k\mathbf{b} - k\mathbf{a}$

3. Compare coefficients of non-parallel vectors to find $\mu$.

Since $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel, we can equate their coefficients on both sides:
Comparing coefficients of $\mathbf{b}$:
$k = 5$
Comparing coefficients of $\mathbf{a}$:
$-\mu = -k \implies \mu = 5$

4. Use vector addition to find $\overrightarrow{OC}$.

Using the triangle law of vector addition:
$\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}$
$\overrightarrow{OC} = \mathbf{b} + (5\mathbf{b} - 5\mathbf{a})$
$\overrightarrow{OC} = 6\mathbf{b} - 5\mathbf{a}$

1. Deduce $\overrightarrow{OC}$ from the information given in part (i).

Since $A$, $B$, and $C$ are collinear, $\overrightarrow{BC}$ must be a scalar multiple of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.
Comparing $\overrightarrow{BC} = 5\mathbf{b} - \mu\mathbf{a}$ with $k(\mathbf{b} - \mathbf{a})$, we get $k = 5$ and $\mu = 5$.
$\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \mathbf{b} + (5\mathbf{b} - 5\mathbf{a}) = 6\mathbf{b} - 5\mathbf{a}$

2. Find the position vector of $N$.

Since $N$ is the midpoint of $OC$,
$\overrightarrow{ON} = \frac{1}{2}\overrightarrow{OC} = \frac{1}{2}(6\mathbf{b} - 5\mathbf{a}) = 3\mathbf{b} - \frac{5}{2}\mathbf{a}$

3. Express the position vector of $E$ using the collinearity of $A$, $E$, and $N$.

Since $E$ lies on the line segment $AN$, we can express $\overrightarrow{OE}$ using a scalar $p$:
$\overrightarrow{OE} = (1-p)\overrightarrow{OA} + p\overrightarrow{ON}$
$\overrightarrow{OE} = (1-p)\mathbf{a} + p\left(3\mathbf{b} - \frac{5}{2}\mathbf{a}\right)$
$\overrightarrow{OE} = \left(1 - \frac{7}{2}p\right)\mathbf{a} + 3p\mathbf{b}$

4. Use the fact that $E$ lies on $OB$ to solve for $p$ and find $\overrightarrow{OE}$.

Since $E$ lies on the line $OB$, its position vector must be a scalar multiple of $\mathbf{b}$. Hence, the coefficient of $\mathbf{a}$ must be zero:
$1 - \frac{7}{2}p = 0 \implies \frac{7}{2}p = 1 \implies p = \frac{2}{7}$
Substitute $p = \frac{2}{7}$ back into the expression for $\overrightarrow{OE}$:
$\overrightarrow{OE} = 3\left(\frac{2}{7}\right)\mathbf{b} = \frac{6}{7}\mathbf{b}$