Answer: \(\overrightarrow{OE}=\lambda\left(\mathbf c+\frac25\mathbf b\right)\), \(\overrightarrow{OE}=\mu\mathbf b+(2-\mu)\mathbf c\), and \(AE:EB=4:3\).

Write each position or direction vector in terms of the given basis vectors, then compare coefficients to find the required constants or ratio.

Since \(C\) is the midpoint of \(OA\),

\(\overrightarrow{OA}=2\mathbf c.\)

Also,

\(\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB} =\mathbf c+\mathbf b.\)

Since \(CD:DB=2:3\),

\(\overrightarrow{CD}=\frac25\overrightarrow{CB} =\frac25\mathbf b.\)

Therefore

Using \(\overrightarrow{OE}=\lambda\overrightarrow{OD}\),

\(\overrightarrow{OE}=\lambda\left(\mathbf c+\frac25\mathbf b\right).\)

This is one expression for \(\overrightarrow{OE}\).

Now use \(\overrightarrow{AE}=\mu\overrightarrow{AB}\). Since

we have

\(\overrightarrow{AE}=\mu(\mathbf b-\mathbf c).\)

Therefore

So

\(\overrightarrow{OE}=\mu\mathbf b+(2-\mu)\mathbf c.\)

Now equate the two expressions for \(\overrightarrow{OE}\):

\(\lambda\mathbf c+\frac25\lambda\mathbf b =\mu\mathbf b+(2-\mu)\mathbf c.\)

Equating coefficients of \(\mathbf b\) and \(\mathbf c\),

\(\frac25\lambda=\mu\)

and

\(\lambda=2-\mu.\)

Substitute \(\mu=\frac25\lambda\) into the second equation:

\(\lambda=2-\frac25\lambda.\)

So

\(\frac75\lambda=2,\)

giving

\(\lambda=\frac{10}{7}.\)

Then

\(\mu=\frac25\cdot\frac{10}{7}=\frac47.\)

Since \(\overrightarrow{AE}=\mu\overrightarrow{AB}\),

\(AE=\frac47AB\)

and

\(EB=\frac37AB.\)

Therefore

\(AE:EB=4:3.\)