Suppose $X$ and $Y$ follow the same distribution, with same mean. And $Var(X)<Var(Y)$. Then, does $X$ first-order stochastically dominate $Y$? Intuitively, I think this will hold. But I do not know how to prove this more formally.
Using the definition that $A$ first order stochastically dominates $B$ if $P(A \geq x) \geq P(B \geq x)$ for all $x$, and for at least one $x$ we have a strict inequality, the answer is no.
Consider two normal distributions centered at $0$. Then if $Y$ has larger variance than $X$, the survival function's $(1-F(x))$ will cross at $0$. Meaning that neither first order scholastically dominates the other.