Answer: \(\overrightarrow{AB}=\mathbf c-2\mathbf a\), \(\overrightarrow{OD}=\frac83\mathbf a+\frac23\mathbf c\), \(\lambda=\frac12\), \(\mu=\frac34\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Since \(\overrightarrow{OC}=\mathbf c\) and \(\overrightarrow{CB}=2\mathbf a\),

\(\overrightarrow{OB}=\mathbf c+2\mathbf a.\)

Also \(\overrightarrow{OA}=4\mathbf a\), so

\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf c+2\mathbf a-4\mathbf a=\mathbf c-2\mathbf a.\)

Since \(AD:DB=2:1\),

\(\overrightarrow{AD}=\frac23\overrightarrow{AB}=\frac23(\mathbf c-2\mathbf a).\)

Therefore

\(\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{AD}=4\mathbf a+\frac23(\mathbf c-2\mathbf a)=\frac83\mathbf a+\frac23\mathbf c.\)

Since \(\overrightarrow{OX}=\mu\overrightarrow{OD}\),

\(\overrightarrow{OX}=\mu\left(\frac83\mathbf a+\frac23\mathbf c\right).\)

Also \(\overrightarrow{AC}=\mathbf c-4\mathbf a\), so

\(\overrightarrow{AX}=\lambda(\mathbf c-4\mathbf a).\)

Now \(\overrightarrow{OX}=\overrightarrow{OA}+\overrightarrow{AX}\), so

\(\mu\left(\frac83\mathbf a+\frac23\mathbf c\right)=4\mathbf a+\lambda(\mathbf c-4\mathbf a).\)

Equating coefficients of \(\mathbf a\) and \(\mathbf c\) gives

\(\frac83\mu=4-4\lambda,\quad \frac23\mu=\lambda.\)

From the second equation, \(\mu=\frac32\lambda\). Substitute into the first:

\(4\lambda=4-4\lambda\), so \(\lambda=\frac12\).

Then \(\mu=\frac34\).

This gives the required answer: \(\overrightarrow{AB}=\mathbf c-2\mathbf a\), \(\overrightarrow{OD}=\frac83\mathbf a+\frac23\mathbf c\), \(\lambda=\frac12\), \(\mu=\frac34\).