1) Lets consider value of n as 1 we get a.
if we consider value of n as 2 we get aaaa.
so language accepted by this language is L = {$a^{1},a^{4},a^{27},a^{256},a^{3125}....$}.
As we can see we can not derive a pattern in this language which we can pump on a state, so this is not regular
2) Lets consider m = 1 and n = 1, we get a.
similarly consider m = 1 and n = 2, we get aa.
with the combination of even and odd string we can derive any string possible, so this language is L = {$a^{+}$}, which is certainly a regular
Now your doubt what happen if we fix the value of m and n in second, it will turn to first problem i.e its not regular.
But in second question we are treating m an n as two different integer, which in turn become a bigger set i.e L = {$a^{+}$}, it already contains all the substring that is in the language L={$(a^{n})^{n}$} and since we are accepting a bigger set, it will accept all its subset.
