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.
Suppose $x^2\equiv -1\pmod{p}$. Then $x^4\equiv 1\pmod{p}$. Since $p = 4k+3$, we have
$$x^{p-1} = x^{4k+2} = x^2x^{4k} \equiv -1(x^4)^k\equiv -1\pmod{p},$$
which contradicts Fermat's Little Theorem.
Best Answer
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, $$ a^{12}\equiv1\pmod{35}. $$