We know that $n^{\text{th}}$ term of the GP $:a_{n} = a\times r^{n-1},$ where $a = $ first term, $r = $ common ratio
Now, $T_{4} = ar^{3} = 25\qquad \rightarrow(1)$
and $T_{6} = ar^{5} = 64\qquad \rightarrow(2)$
Dividing equation $(2)$ by equation $(1),$ we get
$\dfrac{ar^{5}}{ar^{3}} = \dfrac{64}{25}$
$\implies r^{2} = \left(\dfrac{8}{5} \right)^{2}$
$\implies r = \pm \dfrac{8}{5}.$
So, the correct answer is $(A).$