Hint $\:$ Show $ \: x\to x^3\: $ is a bijection via $\rm\color{#c00}{little\ Fermat}$ and $\, \overbrace{3 (2K\!+\!1) = 1 + 2(3K\!+\!1)}^{\textstyle 3J\ \equiv\ 1\ \pmod{p-1}}$
In detail: $ \ \ x^{3J} =(x^{\color{#0a0}{2K+1}})^{\large 3}=\ x (\color{#c00}{x^{3K+1}})^{\large 2} \equiv x\pmod{\!p}\ \ $ for $ \ x\not\equiv 0,\, $ prime $\,p = 3K\!+\!2$.
Thus $ \ x\to x^3\ $ is onto on the finite set $ \:\mathbb Z/p\:,\:$ so it is also $\,1$-$1,\,$ i.e. $ \ x^3 \equiv y^3\, \Rightarrow\, x\equiv y$.
Note: this answers the original version of your question (existence and uniqueness of cube roots).
Remark $ $ the exponent $\,J = \color{#0a0}{2K\!+\!1}$ with $\,x^{3J}\equiv x^{\large 1}\pmod{p=3K\!+\!2}\,$ was computed via
$\!\bmod p\!-\!1=3K\!+\!1\!:\ \ 3J\equiv 1\iff J\equiv \dfrac{1}{3}\equiv \dfrac{-3K}3\equiv -K\equiv \color{#0a0}{2K+1}$
using modular order reduction and $\bmod p\!:\ x^{\large p-1}\equiv 1,\ x\not\equiv 0,\,$ by little Fermat.
Since $\gcd(a,7) =\gcd(a,5) = 1$, from Fermat's theorem,
$$a^6\equiv 1\pmod7 \quad \text{ and } \quad a^4\equiv1\pmod5.
$$
Hence,
$$ a^{12}\equiv 1\pmod7 \quad \text{ and } \quad a^{12}\equiv1\pmod5.
$$
This means that
$$7\mid a^{12}-1 \quad\text{ and } \quad 5\mid a^{12}-1.
$$
Since $\gcd(7,5)=1$,
$$35\mid a^{12}-1,
$$
that is,
$$
a^{12}\equiv1\pmod{35}.
$$
Best Answer
$p^{q-1}+q^{p-1}\equiv p^{q-1}\pmod q$
Now using Fermat's Little Theorem $p^{q-1}\equiv1\pmod q\implies p^{q-1}+q^{p-1}\equiv 1\pmod q$
Similarly, $p^{q-1}+q^{p-1}\equiv 1\pmod p$
As $p$ and $q$ both divides $(p^{q-1}+q^{p-1}-1),$ lcm$(p,q)$ will divide $(p^{q-1}+q^{p-1}-1)$
Now, lcm$(p,q)=p\cdot q$
Generalization:
For any two distinct integer $m,n>1$ where $(m,n)=1,$
$(1):m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod {m\cdot n}$ using Euler' Totient theorem
$(2):m^{\lambda(n)}+n^{\lambda(m)}\equiv1\pmod {m\cdot n}$ using Carmichael Function