= Solution
Put $\epsilon=\hbar^2/(2mR^2)$. The two-particle product <energy eigenstates> are
$$
u_{n_1}(\phi_1)u_{n_2}(\phi_2),\qquad E=\epsilon(n_1^2+n_2^2).
$$
Because the particles are distinguishable, $(n_1,n_2)$ and $(n_2,n_1)$ are distinct when $n_1\ne n_2$. Also, the signs of nonzero integers distinguish opposite <angular momenta>. The <quantum degeneracy> is therefore the number of ordered integer pairs at each sum of squares.
The first distinct <energy levels> are $0,\epsilon,2\epsilon,4\epsilon,5\epsilon,\ldots$. This ordering matters: $4\epsilon$ is the third excited level, while $5\epsilon$ is the fourth excited level.
* At $E=0$, only $(0,0)$ occurs: \b[$g_0=1$].
* At the first excited level, $E=\epsilon$, the pairs are $(\pm1,0)$ and $(0,\pm1)$: \b[$g_1=4$].
* At the fourth excited level, $E=5\epsilon$, the pairs are $(\pm1,\pm2)$ and $(\pm2,\pm1)$, with independent signs: \b[$g_4=8$].
* At $E=25\epsilon$, the pairs are $(\pm5,0),(0,\pm5)$, contributing four, and $(\pm3,\pm4),(\pm4,\pm3)$, contributing eight: \b[$g_{25}=12$].
These counts concern <energy eigenspaces>; arbitrary <quantum superpositions> within each one do not create additional independent states.
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