Before proceeding I looked at the concept which is illustrated in https://en.wikipedia.org/wiki/Integral_of_inverse_functions
I approached as $g\left( x \right) = {f^{ - 1}}\left( x \right)$
$g\left( {f\left( x \right)} \right) = {f^{ - 1}}\left( {f\left( x \right)} \right) = x$
$g\left( {f\left( x \right)} \right) = {f^{ - 1}}\left( {\frac{{g\left( {2x} \right)}}{2}} \right) = x$
$\int\limits_1^8 {xf'\left( x \right)dx} = \frac{q}{p}$
$\left[ {x\int {f'\left( x \right)dx} } \right]_1^8 - \int\limits_1^8 {\frac{d}{{dx}}x\left( {\int {f'\left( x \right)dx} } \right)dx} = \frac{q}{p} \Rightarrow \left[ {xf\left( x \right)} \right]_1^8 - \int\limits_1^8 {f\left( x \right)dx} = \frac{q}{p}$
$ \Rightarrow \left[ {8f\left( 8 \right) - f\left( 1 \right)} \right] - \int\limits_1^8 {f\left( x \right)dx} = \frac{q}{p}$
Not able to procced as not able to use the substitution $g(2x)=2f(x)$
From $g(2x)=2f(x)$ and $f(1)=1,$ one easily derives $f(2^n)=2^n$ for all $n\in\Bbb N$ (by induction), so (according to your calculations) $$\frac qp=63-\int_1^8f(x)dx.$$
Thanks to the link you provided, $$\int_2^4f(x)dx=4^2-2^2-\int_2^4g(x)dx$$ $$=12-2\int_1^2g(2t)dt=12-4\int_1^2f(t)dt=12-5=7,$$ and similarly, $$\int_4^8f(x)dx=8^2-4^2-4\int_2^4f(t)dt=48-28=20,$$ and we can conclude: $$\frac qp=63-\frac54-7-20=\frac{139}4$$ and $p+q=139+4=143.$