= Solution
The <canonical commutation relations> are $[x,p_x]=[y,p_y]=i\hbar$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_xp_y)^\dagger=p_yp_x=p_xp_y$ and $(xy)^\dagger=yx=xy$. Thus $J_1$ is Hermitian and represents an <observable> on the usual oscillator domain.
Introduce <annihilation operators> $a_x=\sqrt{m\omega/(2\hbar)}x+ip_x/\sqrt{2m\hbar\omega}$ and similarly $a_y$. Set $N_x=a_x^\dagger a_x$, $N_y=a_y^\dagger a_y$ and $N=N_x+N_y$. The <Schwinger boson representation> gives
$$
H=\hbar\omega(N+1),\quad
J_1=\frac\hbar2(a_x^\dagger a_y+a_y^\dagger a_x),\quad
J_2=\frac\hbar2(N_y-N_x),\quad
J_3=\frac\hbar{2i}(a_x^\dagger a_y-a_y^\dagger a_x).
$$
For $A=a_x^\dagger a_y$ and $B=a_y^\dagger a_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. Therefore
$$
J_1^2+J_3^2=\frac{\hbar^2}{2}(AB+BA),\qquad
J^2=\frac{\hbar^2}{4}(N^2+2N).
$$
The <two-mode oscillator Casimir identity> is
$$
\boxed{H^2=4\omega^2J^2+\hbar^2\omega^2},\qquad
\boxed{\alpha=4\omega^2,\quad\beta=\hbar^2\omega^2}.
$$
Since $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hbar J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a <common eigenstate of two angular-momentum components> must satisfy $J_3\psi=0$, then the other <commutators> force $J_1\psi=J_2\psi=0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator <ground state>.
Using $J^2=\hbar^2j(j+1)$, with $j=0,1/2,1,\ldots$, and selecting the positive <energy> branch gives
$$
\boxed{E_j=\hbar\omega(2j+1),\qquad g_j=2j+1}.
$$
Equivalently $N=2j=0,1,\ldots$, so $E_N=\hbar\omega(N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\ldots,N$.
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