Answer: \(\overrightarrow{OC}=3\mathbf{b}-2\mathbf{a}\), and \(\lambda=\frac{27}{13}\).

Since

\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a},\)

and \(\frac{AB}{AC}=\frac13\), we have

\(\overrightarrow{AC}=3(\mathbf{b}-\mathbf{a}).\)

Therefore

\(\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}=\mathbf{a}+3(\mathbf{b}-\mathbf{a})=3\mathbf{b}-2\mathbf{a}.\)

Given \(OA:AD=2:7\),

\(\overrightarrow{AD}=\frac72\mathbf{a},\)

so

\(\overrightarrow{OD}=\mathbf{a}+\frac72\mathbf{a}=\frac92\mathbf{a}.\)

A point on \(CD\) can be written as

\(\overrightarrow{OE}=\overrightarrow{OD}+\mu(\overrightarrow{OC}-\overrightarrow{OD}).\)

Thus

\(\overrightarrow{OE}=\frac92\mathbf{a}+\mu\left(3\mathbf{b}-2\mathbf{a}-\frac92\mathbf{a}\right).\)

So

\(\overrightarrow{OE}=\left(\frac92-\frac{13}{2}\mu\right)\mathbf{a}+3\mu\mathbf{b}.\)

But \(\overrightarrow{OE}=\lambda\mathbf{b}\), so the coefficient of \(\mathbf{a}\) is zero:

\(\frac92-\frac{13}{2}\mu=0.\)

Hence

\(\mu=\frac{9}{13}.\)

Therefore

\(\lambda=3\mu=\frac{27}{13}.\)