Prove that $ T(n)=5^n+3T(\lfloor n^\frac{2}{5}\rfloor) $ is $O(5^n)$
Assume that $n \ge 4$ (the induction step is only handling $n \ge 10$ anyway), then $n-n^\frac{2}{5} &gt; n - \sqrt{n} \ge 2$ and you have:
$$
\left( 1 + \frac{3c}{5^{n-n^\frac{2}{5}}} \right) \cdot 5^n \ge
\left( 1 + \frac{3c}{5^2} \right) \cdot 5^n =
\left( 1 + \frac{3}{25} c\right) \cdot 5^n.
$$
The latter term is not larger than $c \cdot 5^n$ whenever $1 + \frac{3}{25} c \le c$, i.e., when $c \ge \frac{25}{22}$.
Pick $c$ large enough to satisfy $c \ge \frac{25}{22}$ and the base cases, and you're done.