= Solution
For $|J,J\rangle$, tracing out subsystem 2 gives the <reduced density matrix>
$$
\rho_1=|j_1,j_1\rangle\langle j_1,j_1|,
$$
which is pure, so its <entanglement entropy> is zero.
For $|J,J-1\rangle$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and gives
$$
\rho_1=
\frac{j_1}{J}|j_1,j_1-1\rangle\langle j_1,j_1-1|
+\frac{j_2}{J}|j_1,j_1\rangle\langle j_1,j_1|.
$$
Its nonzero <eigenvalue>[eigenvalues] are $j_1/J$ and $j_2/J$, so the <Von Neumann entropy> is
$$
S=-\frac{j_1}{J}\log\frac{j_1}{J}
-\frac{j_2}{J}\log\frac{j_2}{J}.
$$
When $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log2$, the maximal entropy for a rank-two reduced state.
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