We have $365 = 5 \times 73$. The congruence becomes $x^2 = -1 \mod 5$ and $x^2 = -1 \mod 73$.
We have if $p = 1 \mod 4 \implies x^2 = -1 \mod p$ has exactly $2$ solutions.
Thus $x^2 = -1 \mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \mod 73$ has solutions $y_0,y_1$.
The original solutions satisfies either:
$x = x_0 \mod 5, x = y_0 \mod 73$; $x = x_0 \mod 5, x = y_1 \mod 73$; $x = x_1 \mod 5, x = y_0 \mod 73$ ; $x = x_1 \mod 5, x = y_1 \mod 73$.
For each pair of congruence , $x$ is uniquely determined $\mod 365$ by the Chinese remainder theorem. Hence the original congruence has $4$ solutions.
. To find solutions of $x^2 - 6x - 13 \equiv 0 \mod 127$, we must write $x^2-6x - 13$ as a square of a linear term, plus a constant. It's seen easily that $x^2 - 6x - 13 = (x-3)^2 - 22$. Hence, the congruence is equivalent to $(x-3)^2 \equiv 22 \mod 127$.
Now, we must check if $22$ is a quadratic residue mod $127$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused.
$$
\binom{22}{127} = \color{green}{\binom{11}{127}}\color{red}{\binom{2}{127}} = \color{green}{-\binom{127}{11}} \times \color{red}1
$$
where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $\binom{2}{p}$ is well known by the remainders which $p$ leaves when divided by $8$.
Now, we can do:
$$
-\binom{127}{11} = -\binom{6}{11} = -\color{green}{\binom{2}{11}}\color{red}{ \binom{3}{11}} = -\color{green}{(-1)}\color{red}{(1)} = 1
$$
Again, the colored terms on the LHS and RHS are equal because the quantities $\binom 2p$ and $\binom 3p$ are well known.
Since we have obtained that the Legendre symbol is $1$, this implies the existence of a solution, and therefore two.
The question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $(x-3)^2 \equiv 22 \mod 127$ is that we basically only need to look for squares of the form $127k + 22$ to find a solution, rather than having to substitute values of $x$ into the expression $x^2-6x-13$ each time.
A brute force : the series $127k + 22 $ goes like : $22,149,276,403,530,657,\color{blue}{784},...$
lo and behold, $784 = 28^2$, hence this gives $x = 31$. Now, note that the congruence actually has two solutions, one given by $127 - 28 = 99$. You can check that $9801 = 99^2 = 22 + 127 \times 77$.
Hence, we get two solutions of $x$, namely $x= 31,102$.
Best Answer
Let's work with the congruence $x^2 \equiv -2 \pmod {122}$.
The ring $\mathbb Z_{122}$ is isomorphic to $\mathbb Z_{2} \times \mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x \in \mathbb Z_2$ and $x' \in \mathbb Z_{61}$.
If $\left(-2\over 61\right)=1$ then there are exactly two solutions to $x^2 \equiv -2 \pmod{61}$. The reason for that is because $\mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 \equiv -2 \pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.
With the congruence $x^2 \equiv 2 \pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $\mathbb Z_{61}$, showing that no solution is possible.