Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? Suppose $X_1\sim N(\mu_1,\sigma^2)$ and $X_2\sim N(\mu_2,\sigma^2)$ where $\mu_2\ge \mu_1$.
Since $\mu_2\ge \mu_1$, based on a characterization of stochastic ordering, we can say that
$$P(X_1>c)\le P(X_2> c) \quad \text{ for any constant }c\,.$$
Now suppose $|\mu_2| \ge |\mu_1|$ instead of $\mu_2 \ge \mu_1$.
Can we also say that $$P(|X_1|>k)\le P(|X_2|> k)\quad \text{ for any constant }k\,?$$
This is certainly the case if $E|X_2|\ge E|X_1|$, but does this hold?
$|X_i|$ has a folded normal distribution and its mean is given by
$$E|X_i|=\sigma\sqrt{\frac 2{\pi}}\exp\left\{-\frac{\mu_i^2}{2\sigma^2}\right\}+\mu_i\left(1-2\Phi\left(-\frac{\mu_i}{\sigma}\right)\right)$$
But it is not immediately clear to me if $E|X_2|\ge E|X_1|$. Is there any simpler way to approach this?

You can reduce your question to the case you understand. Let $s_i$ be the sign of $\mu_i$, that is, $s_i \in \{-1,1\}$ and $s_i \mu_i = |\mu_i|$. Since the question is about the marginals of $(X_1,X_2)$ we can use any coupling between them.
Take $X_1 = \mu_1 + Z$ and $X_2 = \mu_2 + s_1 Z$ where $Z \sim N(0,\sigma^2)$. These will have the correct marginals. Let $X'_1 = |\mu_1| + s_1 Z$ and $X'_2 = |\mu_2| + s_1 s_2 Z$.
\begin{align*}
\mathbb P(|X_1| > k ) = \mathbb P(|s_1 X_1 | > k) &= \mathbb P(|s_1 \mu_1 +s_1 Z| > k) \\
&= \mathbb P(|X'_1| > k) \\
&\le \mathbb P(|X'_2| > k) \\
&= \mathbb P(|s_2 \mu_2 + s_2 s_1 Z| > k) \\
&= \mathbb P(|\mu_2 + s_1 Z| > k) \\
&= \mathbb P(|X_2| > k).
\end{align*}
