$^{95} ≡ 2\; (mod\; 119)$
I know that $119 = 17.7$ and therefore
$^{95} ≡ 2\; (mod\; 7)$ and $^{95} ≡ 2\; (mod\; 17)$.
By Euler Theorem, we have that $x^{\phi(m)} ≡ 1\; (mod\: m)$, when $gcd(x,m) = 1$.
Since $\phi(119) = \phi(17).\phi(7) =16.6 = 96$, we have that
$x^{96} ≡ 1 \; (mod\;119).$
You have that $x^{96} \equiv 1 \pmod {119}$. So since $x^{95} \equiv 2$ we obtain $1 \equiv x^{96} \equiv 2x$, which yields $x = 60$.