This is an answer to the second question, for the particular choice of $\Omega$. In fact, in order to make the notation slightly simpler, let us assume that $\Omega = \{(x_1, x_2) : |x_1 - x_2| > \alpha\}$, with a strict inequality. By a simple approximation argument one easily sees that this change does not influence the answer.

Obviously, if $|m_1 - m_2| \leqslant \alpha$, then $W_c(\mu_1, \mu_2) = 0$: simply take $\gamma$ to be the distribution of $(X + m_1, X + m_2)$, where $X$ has Gaussian distribution with mean $0$ and variance $\sigma^2$.

Suppose that $|m_1 - m_2| > \alpha$, and assume with no loss of generality that $m_2 > m_1$. Let $f_m(x)$ be the probability density function of a Gaussian distribution with mean $m$ and variance $\sigma^2$. Define $\gamma$ in the following way (see the edit at the end of this answer for some intuition):
$$
\begin{aligned} \gamma(dx_1, dx_2) & = \Bigl(\min\{f_{m_1}(x_1), f_{m_2 - \alpha}(x_1)\} \delta_{x_1 + \alpha}(dx_2) \\ & \qquad + \max\{0, f_{m_1}(x_1) - f_{m_2 - \alpha}(x_1)\} \delta_{m_1 + m_2 - x_1}(dx_2)\Bigr) dx_1 . \end{aligned}
$$
This is a distribution concentrated on the line $$x_2 = x_1 + \alpha$$ and the half-line $$\begin{cases} x_2 = m_1 + m_2 - x_1, \\ x_2 \geqslant x_1 + \alpha. \end{cases}$$ It is immediate to see that $\gamma(dx_1, \mathbb{R}) = f_{m_1}(x_1) dx_1$, and it takes a while to find that
$$
\begin{aligned} & \gamma(dx_1, dx_2) \\ & = \Bigl(\min\{f_{m_1}(x_1), f_{m_2 - \alpha}(x_1)\} \delta_{x_2 - \alpha}(dx_1) \\ & \qquad + \max\{0, f_{m_1}(x_1) - f_{m_2 - \alpha}(x_1)\} \delta_{m_1 + m_2 - x_2}(dx_1)\Bigr) dx_2 \\ & = \Bigl(\min\{f_{m_1}(x_2 - \alpha), f_{m_2 - \alpha}(x_2 - \alpha)\} \delta_{x_2 - \alpha}(dx_1) \\ & \qquad + \max\{0, f_{m_1}(m_1 + m_2 - x_2) - f_{m_2 - \alpha}(m_1 + m_2 - x_2)\} \delta_{m_1 + m_2 - x_2}(dx_1)\Bigr) dx_2 \\ & = \Bigl(\min\{f_{m_1 + \alpha}(x_2), f_{m_2}(x_2)\} \delta_{x_2 - \alpha}(dx_1) \\ & \qquad + \max\{0, f_{m_2}(x_2) - f_{m_1 + \alpha}(x_2)\} \delta_{m_1 + m_2 - \alpha - x_2}(dx_1)\Bigr) dx_2\end{aligned}
$$
so that $\gamma(\mathbb{R}, dx_2) = f_{m_2}(x_2) dx_2$. It follows that
$$
\begin{aligned} W_c(\mu_1, \mu_2) & \leqslant \int_{-\infty}^\infty \max\{0, f_{m_1}(x_1) - f_{m_2 - \alpha}(x_1)\} dx_1 \\ & = \int_{-\infty}^{(m_1 + m_2 - \alpha) / 2} (f_{m_1}(x_1) - f_{m_2 - \alpha}(x_1)) dx_1 \\ & = \mathbb{P}(X < \tfrac{m_2 - m_1 - \alpha}{2}) - \mathbb{P}(X < -\tfrac{m_2 - m_1 - \alpha}{2}) \\ & = \mathbb{P}(|X| < \tfrac{m_2 - m_1 - \alpha}{2}) , \end{aligned}
$$
where $X$ has Gaussian distribution with mean $0$ and variance $\sigma^2$.

We claim that the above bound is optimal. Indeed, suppose that $(X_1, X_2)$ is a vector with marginals $f_{m_1}(x_1) dx_1$ and $f_{m_2}(x_2) dx_2$. Then
$$
\begin{aligned}
\mathbb{P}(|X_2 - X_1| > \alpha) & \geqslant \mathbb{P}(X_2 - X_1 > \alpha) \\ & \geqslant \mathbb{P}(X_2 > \tfrac{m_1 + m_2 + \alpha}{2}, \, X_1 < \tfrac{m_1 + m_2 - \alpha}{2}) \\ & \geqslant \mathbb{P}(X_2 > \tfrac{m_1 + m_2 + \alpha}{2}) - \mathbb{P}(X_1 \geqslant \tfrac{m_1 + m_2 - \alpha}{2}) \\ & = \mathbb{P}(X > \tfrac{m_1 - m_2 + \alpha}{2}) - \mathbb{P}(X \geqslant \tfrac{-m_1 + m_2 - \alpha}{2}) \\ & = \mathbb{P}(|X| < \tfrac{m_2 - m_1 - \alpha}{2}) ,
\end{aligned}
$$
with $X$ as above. Thus,
$$
W_c(\mu_1, \mu_2) \geqslant \mathbb{P}(|X| < \tfrac{m_2 - m_1 - \alpha}{2}) .
$$

This gives the desired expression for $W_c(\mu_1, \mu_2)$:
$$
W_c(\mu_1, \mu_2) = \mathbb{P}(|X| < \tfrac{m_2 - m_1 - \alpha}{2}) = 2 \Phi(\tfrac{m_2 - m_1 - \alpha}{2 \sigma}) - 1 ,
$$
where $\Phi$ is the cumulative distribution function of the standard Gaussian distribution.

Edit: The idea behind the coupling $\gamma$ constructed above is in fact quite simple. We like to have $X_1 = X_2 - \alpha$ with as high probability as possible. The "probability" that $X_1 = x_1$ must equal to $f_{m_1}(x_1)$, while the "probability" that $X_2 - \alpha = x_1$ must be equal to $f_{m_2 - \alpha}(x_1)$. Thus, we set $X_1 = X_2 - \alpha = x_1$ with "probability" $\min\{f_{m_1}(x_1), f_{m_2 - \alpha}(x_1)\}$.

This is now extended to a full coupling in a completely arbitrary way. It is quite easy to see that this is indeed possible (any sub-probability distribution $\gamma_0$ with marginals less than $f_{m_1}(x_1) dx_1$ and $f_{m_2}(x_2) dx_2$ can be extended to a full coupling of these measures), and in fact one easily finds an explicit expression by exploiting symmetry of Gaussians.