$5$th term from the end in the expansion of ${\left(\frac{x^2}{3}-\frac{3}{x^2}\right)}^8$is:

As number of terms in the binomial expansion of ${(x+a)}^n$ is $(n+1),$ the $r$th term from the end is $\left(\right(n+1)+(-1\left)\right(r-1\left)\right)th=(n+2-r)\text{ th }$term from the beginning.
Here $n=8, r=5,$ thus, $5$th term from the end is $(8+2-5)$th$5th$term from the beginning, and it is given by
$t_5{=}^8{C}_4{\left(\frac{x^2}{3}\right)}^4{\left(-\frac{3}{x^2}\right)}^{8-4}{=}^8{C}_4=70$