Answer: \(\lambda=\frac4{19}\), \(\mu=\frac6{19}\), \(OX:OQ=6:19\), \(PX:XS=4:15\).

Answer: \(\lambda=\frac4{19}\), \(\mu=\frac6{19}\), \(OX:OQ=6:19\), \(PX:XS=4:15\).

(a) Since \(OP=2\mathbf a\) and \(OS=3\mathbf b\),

\(\overrightarrow{PS}=\overrightarrow{OS}-\overrightarrow{OP}=3\mathbf b-2\mathbf a.\)

Also, \(PX=\lambda PS\), so

\(\overrightarrow{PX}=\lambda(3\mathbf b-2\mathbf a).\)

Therefore

\(\overrightarrow{OX}=\overrightarrow{OP}+\overrightarrow{PX}.\)

Thus

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

So

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

(b) Since \(SR=5\mathbf a\),

\(\overrightarrow{OR}=\overrightarrow{OS}+\overrightarrow{SR}=3\mathbf b+5\mathbf a.\)

Also \(QR=\mathbf b\), so

\(\overrightarrow{OQ}=\overrightarrow{OR}-\overrightarrow{QR}=5\mathbf a+2\mathbf b.\)

(c) Since \(OX=\mu OQ\),

\(\overrightarrow{OX}=\mu(5\mathbf a+2\mathbf b)=5\mu\mathbf a+2\mu\mathbf b.\)

Compare this with

\((2-2\lambda)\mathbf a+3\lambda\mathbf b.\)

Equating coefficients gives

\(2-2\lambda=5\mu,\)

and

\(3\lambda=2\mu.\)

From \(3\lambda=2\mu\),

\(\mu=\frac32\lambda.\)

Substitute into \(2-2\lambda=5\mu\):

\(2-2\lambda=5\left(\frac32\lambda\right).\)

So

\(2=\frac{19}{2}\lambda,\qquad \lambda=\frac4{19}.\)

Then

(d) Since \(OX=\mu OQ\),

\(OX:OQ=\frac6{19}:1=6:19.\)

(e) Since \(PX=\lambda PS\), the remaining fraction is

\(XS=(1-\lambda)PS.\)

Thus

\(PX:XS=\lambda:(1-\lambda)=\frac4{19}:\frac{15}{19}=4:15.\)