$p$ and $q$ are prime, $p \neq q , a$ is integer:
1)$p^{q-1}+q^{p-1} \equiv 1 \pmod{pq}$
2)$p|(a^p + (p-1)!a) $
$p$ and $q$ are prime, $p \neq q , a$ is integer:
1)$p^{q-1}+q^{p-1} \equiv 1 \pmod{pq}$
2)$p|(a^p + (p-1)!a) $
(1)Using Fermat's Little Theorem, $p\mid (q^{p-1}-1)$ as $(p,q)=1$
$\implies p\mid (p^{q-1}+q^{p-1}-1)$
Similarly, $q\mid (p^{q-1}+q^{p-1}-1)\implies $ lcm$(p,q)\mid (p^{q-1}+q^{p-1}-1)$
and the lcm$(p,q)=pq$ as $(p,q)=1$
(2) By Wilson's Theorem, $(p-1)!\equiv-1\pmod p$
So, $a^p+(p-a)!a\equiv a^p-a\pmod p$, but $a^p\equiv a\pmod p$ for all integer $a$ using Fermat's Little Theorem
Hints
1) Work separately modulo $p$ and $q$, using Fermat's Theorem in each case.
2) Use Fermat's Theorem to conclude that $a^p\equiv a\pmod{p}$, then use Wilson's Theorem.
Hint $ $ By Fermat $\rm\: p\,\color{#C00}q\mid p^{q-1} \color{#C00}{q^{p-1}}\!-(\color{#C00}{p^{q-1}\!-1})({q^{p-1}\!-1})\, =\, p^{q-1}\!+q^{p-1}\!-1$
and, secondly, $\rm\,\ p\mid \color{#C00}{a^p-a} + a(\color{#0A0}{1\!+(p\!-\!1)!})\ \,$ by $\,\rm\color{#C00}{Fermat},\, \color{#0A0}{Wilson},\,$ resp.