Finding the probability X > Y I am curious on how to find the probability a random variable X is greater than Y (given their means and variance, and also that they are independent.)
I stumbled across this post: probability of one random variable being greater than another
but am unsure how to answer this question with not necessarily normal distributions.

To get a general answer, knowing just the mean, variance, and that $X$ and $Y$ are independent will not suffice to know this probability. Consider the following example. If $X,Y$ are both $\mathcal{N}(0,1)$, so mean 0 and variance 1, then $P(X>Y) = \frac12$. On the other hand, suppose $X$ and $Y$ are both distributed to have equal probability of being $\pm 1$. Such an $X,Y$ pair will also have mean 0 and variance 1, but now, $P(X>Y) = P(X=1,Y=-1)= \frac14$ (otherwise, it is either the case that $X=Y$ or $-1 = X < Y = 1$).
The general formula for calculating $P(X>Y)$ when $X$ and $Y$ are independent will depend on the respective distributions $F_X$ and $F_Y$ of $X$ and $Y$. A number of equivalent formulae are
$$\begin{aligned}\int P(Y < x)\,\mathrm dF_X(x) &= \int_{-\infty}^x\int_{-\infty}^\infty \mathrm dF_X(x)\mathrm dF_Y(y) \\&= \int_y^\infty \int_{-\infty}^\infty \mathrm dF_Y(y)\mathrm dF_X(x) = \int P(X>y)\,\mathrm dF_Y(y)\end{aligned}$$
Clearly, you will not get an analytic solution except in very special cases like when $Y$ and $X$ have nice formulas for their CDFs like in the normal case.
