= Solution
Label the identical <strange quarks> as 1 and 2 and the <up quark> as 3. In the ground spatial state, the $ss$ pair has symmetric space and flavour factors but an antisymmetric colour factor. The <Pauli exclusion principle> therefore requires a symmetric spin factor: the pair has $S_{ss}=1$.
The <angular momentum> identity
$$
\mathbf S_{ss}^2=(\mathbf S_1+\mathbf S_2)^2
=\mathbf S_1^2+\mathbf S_2^2+2\mathbf S_1\cdot\mathbf S_2
$$
gives
$$
\langle\mathbf S_1\cdot\mathbf S_2\rangle
=\frac12\left[1(1+1)-\frac34-\frac34\right]=\frac14.
$$
Next couple $\mathbf S_{ss}$ to $\mathbf S_u$, where $S_u=1/2$. For total $J$,
$$
\langle\mathbf S_u\cdot\mathbf S_{ss}\rangle
=\frac12\left[J(J+1)-S_u(S_u+1)-S_{ss}(S_{ss}+1)\right].
$$
For the spin-$1/2$ <Xi baryon>, this is $(3/4-3/4-2)/2=-1$. By symmetry the two $us$ pair correlations are each $-1/2$. Inserting the pair correlations into the <baryon hyperfine mass formula> gives
$$
\boxed{M_{uss,J=1/2}=m_u+2m_s+
A'\left(\frac1{4m_s^2}-\frac1{m_um_s}\right).}
$$
For the spin-$3/2$ state, the same coupling gives $(15/4-3/4-2)/2=1/2$. Each $us$ pair correlation is therefore $1/4$, as expected for a fully symmetric spin state. Hence
$$
\boxed{M_{uss,J=3/2}=m_u+2m_s+
A'\left(\frac1{4m_s^2}+\frac1{2m_um_s}\right).}
$$
The predicted splitting is a useful independent check:
$$
M_{uss,J=3/2}-M_{uss,J=1/2}=\frac{3A'}{2m_um_s}>0.
$$
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