Answer: \(\overrightarrow{OQ}=2\mathbf{b}+\mathbf{a}\), \(\overrightarrow{QS}=2\mathbf{a}-2\mathbf{b}\), \(\lambda=\frac34\), \(\mu=\frac58\), \(\frac{QX}{XS}=\frac53\), \(\frac{OR}{OX}=\frac43\).

(a) Since

\(\overrightarrow{OP}=2\mathbf{b}\)

and

\(\overrightarrow{PQ}=\mathbf{a},\)

we have

\(\overrightarrow{OQ}=2\mathbf{b}+\mathbf{a}.\)

(b) Now

\(\overrightarrow{QS}=\overrightarrow{OS}-\overrightarrow{OQ}.\)

Therefore

\(\overrightarrow{QS}=3\mathbf{a}-(2\mathbf{b}+\mathbf{a}) =2\mathbf{a}-2\mathbf{b}.\)

(c) Since \(\overrightarrow{QX}=\mu\overrightarrow{QS}\),

\(\overrightarrow{QX}=\mu(2\mathbf{a}-2\mathbf{b}).\)

Hence

\(\overrightarrow{OX}=\overrightarrow{OQ}+\overrightarrow{QX} =2\mathbf{b}+\mathbf{a}+\mu(2\mathbf{a}-2\mathbf{b}).\)

(d) First,

\(\overrightarrow{OR}=\overrightarrow{OS}+\overrightarrow{SR} =3\mathbf{a}+\mathbf{b}.\)

Since \(\overrightarrow{OX}=\lambda\overrightarrow{OR}\),

\(\overrightarrow{OX}=\lambda(3\mathbf{a}+\mathbf{b}).\)

(e) Equate the two expressions for \(\overrightarrow{OX}\):

\(2\mathbf{b}+\mathbf{a}+\mu(2\mathbf{a}-2\mathbf{b}) =\lambda(3\mathbf{a}+\mathbf{b}).\)

Compare coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):

\(1+2\mu=3\lambda,\)

and

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

Substitute \(\lambda=2-2\mu\) into the first equation:

\(1+2\mu=3(2-2\mu).\)

So

\(1+2\mu=6-6\mu.\)

Hence

\(8\mu=5,\)

so

\(\mu=\frac58.\)

Then

\(\lambda=2-2\left(\frac58\right)=\frac34.\)

(f) Since \(\mu=\frac58\),

\(QX=\frac58QS\)

and

\(XS=\frac38QS.\)

Therefore

\(\frac{QX}{XS}=\frac{5/8}{3/8}=\frac53.\)

(g) Since \(\overrightarrow{OX}=\lambda\overrightarrow{OR}\) and \(\lambda=\frac34\),

\(\frac{OR}{OX}=\frac{1}{\lambda}=\frac43.\)