$\left \{ a^{ \left\lfloor \frac{m}{1} \right\rfloor} :n=1,m>=2 \right \}=\left \{ a^x:x>=2 \right \}$

So, $L_1=\left \{ a^x:x>=2 \right \}$

So, $L_1$ is regular

$L_2=\left \{ a^{ m^1} :n=1,m>=2 \right \}\bigcup \left \{ a^{ m^n} :n<m,n,m>=1 \right \}$

$\left \{ a^{ m^1} :n=1,m>=2 \right \}=\left \{a^x:x>=2\right \}$

So, $L_1=\left \{ a^x:x>=2 \right \}$

So, $L_2$ is regular