Problem 1

If $p=3$, then $a\ge3$, but $a=3$ gives $a!-26<0$, impossible because $m^o>0$; hence $a\ge4$. Direct computation gives $a=4\Rightarrow m^o=4$, so $(m,o)=(2,2)$; and $a=5\Rightarrow m^o=100$, so $(m,o)=(10,2)$. For $a\ge6$, $16\mid a!$, so $m^o\equiv-20\equiv12\pmod{16}$, forcing $m$ even; if $4\mid m$ or $o\ge3$ then $8\mid m^o$, contradicting $m^o\equiv4\pmod8$; and if $m\equiv2\pmod4$ with $o=2$, then $m^2\equiv4\pmod{16}$, so $m^2+20\equiv8\pmod{16}\ne0\equiv a!\pmod{16}$. Thus $a\ge6$ is impossible, leaving exactly the two solutions above.