Answer: \(\boxed{p=1}\), \(\boxed{q=2}\), and \(\boxed{r=-0.646}\) to 3 significant figures.

(i) First find the useful sums.

\[\sum x=1+2+3+4+5=15,\qquad \sum x^2=1^2+2^2+3^2+4^2+5^2=55.\]

Also

\[\sum y=4+p+q+2+1=7+p+q,\]

and

\[\sum xy=1(4)+2p+3q+4(2)+5(1)=17+2p+3q.\]

Now

\[S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=55-\frac{15^2}{5}=10.\]

Also

\[S_{xy}=\sum xy-\frac{(\sum x)(\sum y)}{n}.\]

So

\[S_{xy}=17+2p+3q-\frac{15(7+p+q)}{5}.\]

This simplifies to

\[S_{xy}=-4-p.\]

The gradient of the regression line of \(y\) on \(x\) is

\[\frac{S_{xy}}{S_{xx}}.\]

The given regression line is \(y=-0.5x+3.5\), so

\[-0.5=\frac{-4-p}{10}.\]

Hence

\[-5=-4-p,\]

so

\[\boxed{p=1}.\]

The regression line passes through \((\bar{x},\bar{y})\). Since \(\bar{x}=15/5=3\),

\[\bar{y}=-0.5(3)+3.5=2.\]

Therefore

\[\frac{7+p+q}{5}=2.\]

Using \(p=1\),

\[7+1+q=10,\]

so

\[\boxed{q=2}.\]

(ii) Now the \(y\)-values are \(4,1,2,2,1\). Hence

\[\sum y=10,\qquad \sum y^2=4^2+1^2+2^2+2^2+1^2=26.\]

So

\[S_{yy}=26-\frac{10^2}{5}=6.\]

Also, using \(p=1\),

\[S_{xy}=-4-p=-5.\]

The product moment correlation coefficient is

\[r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}.\]

Therefore

\[r=\frac{-5}{\sqrt{10\cdot6}}=-0.6455\ldots.\]

So

\[\boxed{r=-0.646}\quad\text{to 3 significant figures}.\]