Problem 1

If $p=1$, then $a!=m^o+25$, and $a!\ge26$ forces $a\ge5$, so $5\mid m$. If $o$ is even then $m^o\equiv1\pmod4$ (as $m$ is odd), giving $a!\equiv2\pmod4$, impossible since $a\ge5$; hence $o$ is odd, $o\ge3$. Since $5^3\mid m^o$, we get $a!\equiv25\pmod{125}$, which forces $10\le a\le14$; a direct check shows $5^3\nmid a!-25$ for each such $a$, so $p=1$ gives no solution.