Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.

For any prime $p$, you can show that $x^2\equiv1\bmod p\iff x\equiv \pm1\bmod p$. To this end, notice that $$x^2\equiv 1\bmod p\iff \exists k\in\mathbb Z: x^2=k\cdot p+1\iff (x+1)(x-1)=k\cdot p$$ Euclid's Lemma implies now that either $p\mid x+1\iff x\equiv -1\bmod p$ or $p\mid x-1\iff x\equiv 1\bmod p$.