Answer: \(\overrightarrow{AB}=\mathbf b-\mathbf a\), \(\overrightarrow{PQ}=\frac12\mathbf b-\frac14\mathbf a\), \(n=2\), \(k=\frac12\).

(a) Since \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OB}=\mathbf b\),

\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} =\mathbf b-\mathbf a.\)

(b) The position vector of \(P\) is

\(\overrightarrow{OP}=\frac34\mathbf a.\)

Since \(Q\) is the midpoint of \(AB\),

\(\overrightarrow{OQ}=\frac12(\mathbf a+\mathbf b).\)

Therefore

\(\overrightarrow{PQ} =\overrightarrow{OQ}-\overrightarrow{OP} =\frac12(\mathbf a+\mathbf b)-\frac34\mathbf a.\)

So

\(\overrightarrow{PQ} =\frac12\mathbf b-\frac14\mathbf a.\)

(c) Since \(n\overrightarrow{PQ}=\overrightarrow{QR}\),

\(\overrightarrow{QR} =n\left(\frac12\mathbf b-\frac14\mathbf a\right).\)

(d) Since \(\overrightarrow{BR}=k\mathbf b\),

\(\overrightarrow{OR} =\overrightarrow{OB}+\overrightarrow{BR} =\mathbf b+k\mathbf b =(1+k)\mathbf b.\)

Hence

\(\overrightarrow{QR} =\overrightarrow{OR}-\overrightarrow{OQ} =(1+k)\mathbf b-\frac12(\mathbf a+\mathbf b).\)

So

\(\overrightarrow{QR} =-\frac12\mathbf a+\left(\frac12+k\right)\mathbf b.\)

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

\(n\left(\frac12\mathbf b-\frac14\mathbf a\right) =-\frac12\mathbf a+\left(\frac12+k\right)\mathbf b.\)

Compare coefficients of \(\mathbf a\):

\(-\frac n4=-\frac12.\)

Therefore

\(n=2.\)

Compare coefficients of \(\mathbf b\):

\(\frac n2=\frac12+k.\)

Substitute \(n=2\):

\(1=\frac12+k.\)

Therefore

\(k=\frac12.\)