= Solution
For $|w|\gg|\Delta|$, the mixed states approach opposite-spin product states and
$$
\boxed{E_\pm\simeq-\frac\Delta4\pm\frac{|w|}{2},\qquad E_{\uparrow\uparrow,\downarrow\downarrow}=\frac\Delta4\mp\frac u2.}
$$
The omitted mixed-branch correction is of order $\Delta^2/|w|$. The four levels are \b[generically nondegenerate], apart from special moment choices or crossings.
For equal-and-opposite signed moments, $g_2=-g_1$, $u=0$ and the parallel-spin level remains exactly twofold degenerate at $\Delta/4$; the other two approach $-\Delta/4\pm|g_1\mu_0B|$. Equal moment magnitudes are an extra assumption, not implied by opposite charges alone. If $g_1=g_2$, the differential Zeeman term vanishes and the singlet/triplet-zero block stays unmixed by the field. This is the <strong-field limit of a two-spin exchange spectrum>.
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