Answer: \(\boxed{b=1.516}\), \(\boxed{p=4.2}\), and the estimate is \(\boxed{x=4.73}\).

(i) For the two regression lines, the product of their gradients is \(r^2\). Therefore

\[b(0.6331)=0.9797^2.\]

So

\[b=\frac{0.9797^2}{0.6331}=1.516051\ldots.\]

Thus

\[\boxed{b=1.516}\]

to 3 decimal places.

(ii) The regression line of \(y\) on \(x\) passes through \((\bar{x},\bar{y})\). Since the sum of the \(y\)-values is \(46.5\) and there are \(6\) observations,

\[\bar{y}=\frac{46.5}{6}=7.75.\]

Use \(y=bx+1.306\):

\[7.75=b\bar{x}+1.306.\]

Using \(b=1.516051\ldots\),

\[\bar{x}=\frac{7.75-1.306}{1.516051\ldots}=4.2505\ldots.\]

Hence

\[\sum x=6\bar{x}=25.503\ldots.\]

The known \(x\)-values add to

\[2.3+2.8+3.7+6.1+6.4=21.3.\]

Therefore

\[p=25.503\ldots-21.3=4.203\ldots.\]

So

\[\boxed{p=4.2}\]

to 1 decimal place.

(iii) First find \(d\) using the fact that the regression line of \(x\) on \(y\) also passes through \((\bar{x},\bar{y})\):

\[\bar{x}=0.6331\bar{y}+d.\]

So

\[d=4.2505\ldots-0.6331(7.75)=-0.6560\ldots.\]

For \(y=8.5\),

\[x=0.6331(8.5)-0.6560\ldots=4.7253\ldots.\]

Therefore the estimate is

\[\boxed{x=4.73}.\]