= Solution
With $\hbar=1$, <angular momentum eigenstates> have $j=0,\tfrac12,1,\ldots$ and $m=-j,-j+1,\ldots,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\geq1/2$, $J=j-1/2$. For the <Clebsch-Gordan coefficients> to be nonzero one requires $\boxed{M=r+m}$.
For $j=2$, write $|\pm,m\rangle$ for $|1/2,\pm1/2\rangle|2,m\rangle$. The highest state and one application of the <angular momentum lowering operator> give
$$
\boxed{\begin{aligned}
|5/2,5/2\rangle&=|+,2\rangle,\\
|5/2,3/2\rangle&=\frac1{\sqrt5}|-,2\rangle+\frac2{\sqrt5}|+,1\rangle,\\
|3/2,3/2\rangle&=\frac2{\sqrt5}|-,2\rangle-\frac1{\sqrt5}|+,1\rangle.
\end{aligned}}
$$
The lowering coefficient of the highest $J=5/2$ state is $\sqrt5$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\geq3/2$ vanish.
For the decay, the final total intrinsic spin is $1/2$. Combining it with integer <orbital angular momentum> $\ell$ to obtain $J=3/2$ permits only $\ell=1,2$. <Parity conservation> gives $\eta_X=\eta_A\eta_B(-1)^\ell=(-1)^\ell$. Thus \b[$\boxed{\eta_X=+1:\ell=2;\quad\eta_X=-1:\ell=1}$].
For $\ell=2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell=1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2\rangle|1,1\rangle$. Therefore \b[$\boxed{P=1/5\ \hbox{for }\eta_X=+1,\qquad P=1\ \hbox{for }\eta_X=-1}$].
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