To make the computations easy to understand, we first consider general equations.
Let $a,b,n$ be integers.
Suppose gcd$(a, n) = 1$.
Consider the following equation.
$ax \equiv b$ (mod $n$)
Since gcd$(a, n) = 1$, we can solve $ay \equiv 1$ (mod $n$) by Euclid's algorithm.
Then $x = by$ (mod $n$) is the solution.
Let $a,b,n,m$ be integers.
Suppose gcd$(n, m) = 1$.
Consider the following equatiuons.
$x \equiv a$ (mod $n$)
$x \equiv b$ (mod $m$)
Since gcd$(n, m) = 1$, we can find, by Euclid's algorithm, integers $k, l$ such that
$mk \equiv 1$ (mod $n$)
$nl \equiv 1$ (mod $m$)
Then $x = amk + bnl$ (mod $nm$) is the solution.
Now let's solve the given equations
$5k \equiv 3$ (mod $6$)
$5k \equiv 2$ (mod $7$)
We get(by Euclid's algorithm or just by testing)
$k \equiv 3$ (mod $6$)
$k \equiv 6$ (mod $7$)
Then we can apply the above method to find $k$.
Since
$7 \equiv 1$ (mod $6$)
$-6 \equiv 1$ (mod $7$)
$k = 3\cdot7 -6\cdot6 = -15 \equiv 27$ (mod $42$)
$5\nmid (n^2+n+1)$ for all integer $n$. Because
$$n^2+n+1 \equiv (n+3)^2 + 2 \pmod 5$$
So if $5\mid (n^2+n+1)$ for some $n$, then $n$ satisfies $(n+3)^2 \equiv 3 \pmod 5$. But $3$ is not quadratic residue modulo $5$. (You can check that $3$ is not quadratic residue easily.)
Best Answer
As $39=13\cdot3$
For non-negative integers $m,n$
$\displaystyle53\equiv1\pmod{13}\implies53^n\equiv1$ and $\displaystyle103\equiv-1\pmod{13}\implies103^{53}\equiv(-1)^{53}$
$\displaystyle\implies53^{103}+103^{53}\equiv1+(-1)\pmod{13}$
and $\displaystyle53\equiv-1\pmod3\implies53^{103}\equiv(-1)^{103}$ and $\displaystyle103\equiv1\pmod3\implies103^m\equiv1$
$\displaystyle\implies53^{103}+103^{53}\equiv-1+(1)\pmod3$