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With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. 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 $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, 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 $M\ge 3/2$ 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 $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

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 $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.