Solution:
(i) In \[\vartriangle CPQ\text{ }and\text{ }\vartriangle CAB\]
\[\angle PCQ\text{ }=\angle APQ\]
[As PQ || AB, corresponding angles are equal.]
\[\angle C\text{ }=\angle C\]
[Common angle]
Hence, \[\vartriangle CPQ\text{ }\sim\text{ }\vartriangle CAB\text{ }by\text{ }AA\] criterion for similarity
So, we have
\[CP/CA\text{ }=\text{ }CQ/CB\]
\[CP/CA\text{ }=\text{ }4.8/\text{ }8.4\text{ }=\text{ }4/7\]
Thus, \[CP/PA\text{ }=\text{ }4/3\]
(ii) As, \[\vartriangle CPQ\text{ }\sim\text{ }\vartriangle CAB\text{ }by\text{ }AA\] criterion for similarity
We have,
\[PQ/AB\text{ }=\text{ }CQ/CB\]
\[PQ/6.3\text{ }=\text{ }4.8/8.4\]
So,
\[PQ\text{ }=\text{ }3.6\text{ }cm\]