= Solution
For <spin> one-half, $s_\pm=s_x\pm is_y=(\hbar/2)(\sigma_x\pm i\sigma_y)$, hence
$$
\boxed{s_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad s_-=\hbar\begin{pmatrix}0&0\\1&0\end{pmatrix}.}
$$
They raise or lower the <spin> projection by one unit of $\hbar$ and annihilate the corresponding endpoint. The exchange identity $\boldsymbol\sigma_1\cdot\boldsymbol\sigma_2=\sigma_{1z}\sigma_{2z}+2(\sigma_{1+}\sigma_{2-}+\sigma_{1-}\sigma_{2+})$ preserves total <spin> projection.
Write $\Delta=\Delta E$, $u=\mu_0B(g_1+g_2)$ and $w=\mu_0B(g_1-g_2)$. In the \b[listed] order $(\uparrow\uparrow,\downarrow\downarrow,\uparrow\downarrow,\downarrow\uparrow)$, direct action gives
$$
H=\begin{pmatrix}
\Delta/4-u/2&0&0&0\\
0&\Delta/4+u/2&0&0\\
0&0&-\Delta/4-w/2&\Delta/2\\
0&0&\Delta/2&-\Delta/4+w/2
\end{pmatrix}.
$$
\b[The printed matrix instead uses $(\uparrow\uparrow,\uparrow\downarrow,\downarrow\uparrow,\downarrow\downarrow)$.] Permuting the listed indices to $(1,3,4,2)$ recovers its central block and diagonal entries. This is a basis-order mismatch in the source, not a different <Hamiltonian>.
The <two-spin exchange spectrum with unequal signed magnetic moments> is
$$
\boxed{E_{\uparrow\uparrow}=\frac\Delta4-\frac u2,\quad E_{\downarrow\downarrow}=\frac\Delta4+\frac u2,\quad E_\pm=-\frac\Delta4\pm\frac12\sqrt{\Delta^2+w^2}.}
$$
Use the products $g_j\mu_0$ as the signed magnetic coefficients in the stated <Hamiltonian>. Opposite physical charges must be represented in those signs; they do not by themselves imply equal moment magnitudes. Here $\mu_0$ denotes the source's magneton, not vacuum permeability.
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