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As given that 1st is not regular and 2nd is regular as 1st not form AP but 2nd form.but if in 2nd we fix value of m and n same then it will work as 1st(not regular) so 2nd  also should not be regular.as i know if we fix m or n value as any constant it will be AP but what if same???

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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.
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