Answer: (b) \(\overrightarrow{YZ}=\lambda\left(\frac23\mathbf a-\frac7{15}\mathbf b\right)\); (c) \(\overrightarrow{YZ}=\left(\mu+\frac13\right)\mathbf a-\frac13\mathbf b\); (d) \(\lambda=\frac57,\ \mu=\frac17\).

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

(a) First find the position vector of \(Y\).

Since

\(\overrightarrow{AY}=\frac13\overrightarrow{AB},\)

we have

\(\overrightarrow{OY}=\overrightarrow{OA}+\overrightarrow{AY}.\)

Now

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

so

Also

\(\overrightarrow{OX}=\frac45\mathbf b.\)

Therefore

\(\overrightarrow{XY}=\overrightarrow{OY}-\overrightarrow{OX}.\)

Hence

\(\overrightarrow{XY}=\frac23\mathbf a+\frac13\mathbf b-\frac45\mathbf b.\)

So

\(\overrightarrow{XY}=\frac23\mathbf a-\frac7{15}\mathbf b,\)

as required.

(b) Since \(\overrightarrow{YZ}=\lambda\overrightarrow{XY}\),

\(\overrightarrow{YZ}=\lambda\left(\frac23\mathbf a-\frac7{15}\mathbf b\right).\)

(c) Since \(\overrightarrow{AZ}=\mu\mathbf a\),

\(\overrightarrow{OZ}=\overrightarrow{OA}+\overrightarrow{AZ}=(1+\mu)\mathbf a.\)

Therefore

\(\overrightarrow{YZ}=\overrightarrow{OZ}-\overrightarrow{OY}.\)

So

\(\overrightarrow{YZ}=(1+\mu)\mathbf a-\left(\frac23\mathbf a+\frac13\mathbf b\right).\)

Thus

\(\overrightarrow{YZ}=\left(\mu+\frac13\right)\mathbf a-\frac13\mathbf b.\)

(d) Equate the two expressions for \(\overrightarrow{YZ}\):

Comparing the coefficients of \(\mathbf b\),

\(-\frac7{15}\lambda=-\frac13.\)

So

\(\lambda=\frac57.\)

Comparing the coefficients of \(\mathbf a\),

\(\frac23\lambda=\mu+\frac13.\)

Substitute \(\lambda=\frac57\):

\(\frac23\cdot\frac57=\mu+\frac13.\)

So

\(\frac{10}{21}=\mu+\frac13.\)

Therefore

\(\mu=\frac{10}{21}-\frac7{21}=\frac17.\)