CBEO is a parallelogram. The position vectors of C and E, relative to O, are $\mathbf{a}$ and $\mathbf{b}$ respectively.

1. Relate $\\overrightarrow{BD}$ to $\\overrightarrow{BC}$ using the given ratio.

Since $3\\overrightarrow{BC}=5\\overrightarrow{BD}$,
$\\overrightarrow{BD}=\\frac{3}{5}\\overrightarrow{BC}.$

2. Find $\\overrightarrow{CD}$ from the whole vector $\\overrightarrow{CB}$.

In parallelogram $CBEO$, $\\overrightarrow{CB}=\\overrightarrow{OE}=\\mathbf{b}$.

Since $D$ lies on $CB$,
$\\overrightarrow{CD}=\\overrightarrow{CB}-\\overrightarrow{DB}.$
But $\\overrightarrow{DB}=-\\overrightarrow{BD}$ is not needed if using lengths along the same direction from $C$ to $B$:
$\\overrightarrow{CD}=\\overrightarrow{CB}-\\overrightarrow{BD}$
$=\\mathbf{b}-\\frac{3}{5}\\mathbf{b}$
$=\\frac{2}{5}\\mathbf{b}.$

1. Find the position vector of $D$ using $\overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{CD}$.

From part (a)(i),
$\overrightarrow{CD}=\frac{2}{5}\mathbf{b}.$
Hence
$\overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{CD}=\mathbf{a}+\frac{2}{5}\mathbf{b}.$

2. Find $\overrightarrow{DE}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\overrightarrow{DE}=\overrightarrow{OE}-\overrightarrow{OD}$
$=\mathbf{b}-\left(\mathbf{a}+\frac{2}{5}\mathbf{b}\right)$
$=\frac{3}{5}\mathbf{b}-\mathbf{a}.$

3. Use $\overrightarrow{AD}=\frac{2}{3}\overrightarrow{DE}$ to find $\overrightarrow{AD}$.

Given
$AD=\frac{2}{3}DE,$
and $A,D,E$ are collinear with $D$ between $A$ and $E$, so
$\overrightarrow{AD}:\overrightarrow{DE}=2:3.$
Thus
$\overrightarrow{AD}=\frac{2}{3}\overrightarrow{DE}.$
$\overrightarrow{AD}=\frac{2}{3}\left(\frac{3}{5}\mathbf{b}-\mathbf{a}\right)$
$=\frac{2}{5}\mathbf{b}-\frac{2}{3}\mathbf{a}.$

1. Use the earlier result for $\overrightarrow{AD}$ and the position of $D$ to find $\overrightarrow{AC}$.

From part (a)(i),
$\overrightarrow{CD}=\dfrac{2}{5}\mathbf{b}$, so
$\overrightarrow{DC}=-\dfrac{2}{5}\mathbf{b}.$
From part (a)(ii),
$\overrightarrow{AD}=\dfrac{3}{5}\mathbf{b}-\dfrac{3}{5}\mathbf{a}.$
Hence
$\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}$
$=\left(\dfrac{3}{5}\mathbf{b}-\dfrac{3}{5}\mathbf{a}\right)-\dfrac{2}{5}\mathbf{b}$
$=\dfrac{1}{5}\mathbf{b}-\dfrac{3}{5}\mathbf{a}.$

2. Compare $\overrightarrow{AC}$ with $\overrightarrow{OC}$ to decide whether the two lines are parallel.

Since
$\overrightarrow{OC}=\mathbf{a},$
for $AC$ to be parallel to $OC$, $\overrightarrow{AC}$ must be a scalar multiple of $\mathbf{a}$.

But
$\overrightarrow{AC}=\dfrac{1}{5}\mathbf{b}-\dfrac{3}{5}\mathbf{a},$
which contains a non-zero $\mathbf{b}$ term. Therefore $\overrightarrow{AC}$ is not a scalar multiple of $\overrightarrow{OC}$.

1. Compare the areas using the bases on the same straight line $CB$.

Since $D$, $B$ and $C$ lie on the same straight line, $\Delta ADB$ and $\Delta ACD$ have the same perpendicular height from $A$ to line $CB$.

Hence
$\frac{\text{area of }\Delta ADB}{\text{area of }\Delta ACD}=\frac{BD}{CD}.$From part (a)(i), $\overrightarrow{CD}=\dfrac{2}{5}\mathbf{b}$. Also, from $3BC=5BD$, $BD=\dfrac{3}{5}BC$. Since $CB=\mathbf{b}$,
$CD:BD=\frac{2}{5}BC:\frac{3}{5}BC=2:3.$
Therefore
$\frac{\text{area of }\Delta ADB}{\text{area of }\Delta ACD}=\frac{BD}{CD}=\frac{3}{2}.$

1. Relate the areas of $\Delta ADC$ and $\Delta AEC$ using bases on the same straight line $AE$.

Since $A,D,E$ are collinear and $AD=\dfrac{2}{3}DE$, we have
$AD:DE=2:3,$
so
$AD:AE=2:5.$
Triangles $ADC$ and $AEC$ have bases $AD$ and $AE$ on the same straight line $AE$, and share the same height from $C$.

Hence
$\frac{\text{area of }\Delta ADC}{\text{area of }\Delta AEC}=\frac{AD}{AE}=\frac{2}{5}.$

2. Connect $\Delta AEC$ and $\Delta AOE$ using the same base $AE$.

From part (b), $AC$ is parallel to $OE$. Since $O$ and $C$ lie on the same line parallel to $AE$? No direct equality of heights is needed if using the vector result: $\overrightarrow{AC}=-\dfrac{3}{5}\mathbf b$ and $\overrightarrow{OE}=\mathbf b$, so $AC \parallel OE$.

Thus triangles $AEC$ and $AOE$ have the same base $AE$ and equal perpendicular heights, since $C$ and $O$ lie on a line parallel to $AE$.

Therefore
$\text{area of }\Delta AEC=\text{area of }\Delta AOE.$

3. Combine the area ratios to obtain the required ratio.

Therefore
$\frac{\text{area of }\Delta ADC}{\text{area of }\Delta AOE}=\frac{2}{5}.$

1. Express the area of $\Delta ADB$ and $\Delta CDE$ in terms of the area of $\Delta AOE$.

From earlier parts,
$\frac{\text{area of }\Delta ADB}{\text{area of }\Delta ACD}=\frac{3}{2}$
and
$\frac{\text{area of }\Delta ACD}{\text{area of }\Delta AOE}=\frac{2}{5}.$
Hence
$\frac{\text{area of }\Delta ADB}{\text{area of }\Delta AOE}=\frac{3}{2}\times \frac{2}{5}=\frac{3}{5}.$
Also, since $AD:DE=2:3$, triangles $ACD$ and $CDE$ have bases $AD$ and $DE$ on the same straight line $AE$ and the same height from $C$.
$\frac{\text{area of }\Delta CDE}{\text{area of }\Delta ACD}=\frac{DE}{AD}=\frac{3}{2}.$
So
$\frac{\text{area of }\Delta CDE}{\text{area of }\Delta AOE}=\frac{3}{2}\times \frac{2}{5}=\frac{3}{5}.$

2. Find the area of quadrilateral $CDEO$ in terms of $\Delta AOE$, then form the required ratio.

Since $\overrightarrow{AC}$ is parallel to $\overrightarrow{OE}$, triangles $CEO$ and $AEO$ have the same base $OE$ and equal heights.
$\text{area of }\Delta CEO=\text{area of }\Delta AOE.$
Now
$\text{area of }CDEO=\text{area of }\Delta CDE+\text{area of }\Delta CEO.$
Therefore
$\text{area of }CDEO=\frac{3}{5}\text{area of }\Delta AOE+\text{area of }\Delta AOE=\frac{8}{5}\text{area of }\Delta AOE.$
Hence
$\frac{\text{area of }\Delta ADB}{\text{area of }CDEO}=\frac{\frac{3}{5}\text{area of }\Delta AOE}{\frac{8}{5}\text{area of }\Delta AOE}=\frac{3}{8}.$