Answer: \(\overrightarrow{RQ}=\binom{-\frac52}{5}\), \(\alpha=3\), \(\beta=-1\).

From \(4\overrightarrow{PR}=\binom{-2}{8}\),

\(\overrightarrow{PR}=\binom{-\frac12}{2}.\)

Since

\(\overrightarrow{RQ}=\overrightarrow{RP}+\overrightarrow{PQ},\)

we have

\(\overrightarrow{RQ}=-\overrightarrow{PR}+\overrightarrow{PQ} =\binom{\frac12}{-2}+\binom{-3}{7} =\binom{-\frac52}{5}.\)

For the second part,

\(3\mathbf{a}-2\mathbf{b}=3(\alpha\mathbf{i}+6\mathbf{j})-2(4\mathbf{i}+\beta\mathbf{j}).\)

So

\(3\mathbf{a}-2\mathbf{b}=(3\alpha-8)\mathbf{i}+(18-2\beta)\mathbf{j}.\)

Equating this with \(\mathbf{c}=(2\alpha+5\beta)\mathbf{i}+20\mathbf{j}\),

\(3\alpha-8=2\alpha+5\beta,\)

and

\(18-2\beta=20.\)

The second equation gives \(\beta=-1\). Substitute this into the first equation:

\(3\alpha-8=2\alpha-5,\)

so

\(\alpha=3.\)