= Solution
Write $|m_1,m_2\rangle$ for product <spin> states, and $J_+=J_{1+}+J_{2+}$. A highest-weight state is killed by $J_+$. For $s=2$ the unique $m=2$ state is $|1,1\rangle$. In the $m=1$ subspace, $J_+$ kills $a|1,0\rangle+b|0,1\rangle$ exactly when $a+b=0$. In the $m=0$ subspace it kills $a|1,-1\rangle+b|0,0\rangle+c|-1,1\rangle$ exactly when $a+b=b+c=0$. Normalizing gives
$$
\boxed{
\begin{aligned}
|2,2\rangle&=|1,1\rangle,\\
|1,1\rangle&=\frac{|1,0\rangle-|0,1\rangle}{\sqrt2},\\
|0,0\rangle&=\frac{|1,-1\rangle-|0,0\rangle+|-1,1\rangle}{\sqrt3}.
\end{aligned}}
$$
The left states are coupled states and the right states are product states. Lowering commutes with particle interchange, so the whole <spin>-$s$ multiplet has the exchange sign of its highest-weight state: $+1$ for $s=0,2$ and $-1$ for $s=1$, or $(-1)^s$. Integer-<spin> identical particles are <bosons>; their spatial partial wave changes by $(-1)^\ell$ under interchange. Symmetry of the full state therefore requires \b[$\ell+s$ even].
For decay from total <angular momentum> one, coupling requires $|\ell-s|\le1\le\ell+s$. The case $s=0$ forces $\ell=1$ and violates Bose symmetry. For $s=1$, the allowed range is $\ell=0,1,2$ and symmetry leaves $\ell=1$. For $s=2$, the range is $\ell=1,2,3$ and symmetry leaves $\ell=2$. The identical particles' intrinsic <parity> product is $+1$, so total final <parity> is $(-1)^\ell$. Hence
$$
\boxed{P_X=-1:\ (\ell,s)=(1,1),\qquad
P_X=+1:\ (\ell,s)=(2,2).}
$$
This uses the stated massive-<spin> angular-momentum model; no additional restrictions for massless transverse vector particles have been assumed.
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