Answer: (i) \(\dfrac5{14}(7x-10)^{\frac25}+C\); (ii) \(a=\dfrac{253}{7}\).

For part (i), increase the power by \(1\):

\(-\dfrac35+1=\dfrac25.\)

Also divide by the derivative of \(7x-10\), which is \(7\). Therefore

\(\int(7x-10)^{-\frac35}\,dx=\dfrac{(7x-10)^{\frac25}}{7\cdot\frac25}+C.\)

So

\(\int(7x-10)^{-\frac35}\,dx=\dfrac5{14}(7x-10)^{\frac25}+C.\)

For part (ii), use the result from part (i):

\(\dfrac5{14}\left[(7x-10)^{\frac25}\right]_6^a=\dfrac{25}{14}.\)

So

\(\dfrac5{14}(7a-10)^{\frac25}-\dfrac5{14}(7\cdot6-10)^{\frac25}=\dfrac{25}{14}.\)

Since \(7\cdot6-10=32\) and \(32^{\frac25}=4\),

\(\dfrac5{14}(7a-10)^{\frac25}-\dfrac{20}{14}=\dfrac{25}{14}.\)

Hence

\(\dfrac5{14}(7a-10)^{\frac25}=\dfrac{45}{14}.\)

So

\((7a-10)^{\frac25}=9.\)

Raise both sides to the power \(\dfrac52\):

\(7a-10=9^{\frac52}=243.\)

Therefore

\(7a=253,\)

so

\(\boxed{a=\dfrac{253}{7}}.\)