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,

\[PQ\text{ }=\text{ }3.6\text{ }cm\]