Answer: \(\overrightarrow{DE}=(h-1)\mathbf c+\frac34(h-1)\mathbf a\), \(\overrightarrow{DE}=\frac14\mathbf a+k\mathbf c\), \(h=\frac43\), \(k=\frac13\).

Since \(D\) lies on \(CB\) with \(CD:DB=3:1\),

\(\overrightarrow{CD}=\frac34\overrightarrow{CB}.\)

In the parallelogram, \(\overrightarrow{CB}=\mathbf a\). Hence

\(\overrightarrow{OD}=\mathbf c+\frac34\mathbf a.\)

(a) Since

\(\overrightarrow{OE}=h\overrightarrow{OD},\)

we have

Therefore

\(\overrightarrow{DE} =\overrightarrow{OE}-\overrightarrow{OD}.\)

So

Thus

(b) Since \(\overrightarrow{BE}=k\overrightarrow{AB}\) and \(\overrightarrow{AB}=\mathbf c\),

\(\overrightarrow{BE}=k\mathbf c.\)

Also

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

Hence

Therefore

So

\(\overrightarrow{DE}=\frac14\mathbf a+k\mathbf c.\)

(c) Equate the two expressions for \(\overrightarrow{DE}\):

Compare the coefficients of \(\mathbf a\):

\(\frac34(h-1)=\frac14.\)

Thus

\(h-1=\frac13,\)

so

\(h=\frac43.\)

Compare the coefficients of \(\mathbf c\):

\(k=h-1=\frac13.\)