 Solution

On the tensor product state space, the total angular momentum is $\mathbf{J}={\mathbf{J}}^{(1)}\otimes I+I\otimes {\mathbf{J}}^{(2)}$, conventionally abbreviated ${\mathbf{J}}^{(1)}+{\mathbf{J}}^{(2)}$. Operators on different factors commute. Define $J_{\pm }^{(i)}=J_1^{(i)}\pm i{J}_2^{(i)}$. Expanding the square givesThe first three terms commute with $J_3=J_3^{(1)}+J_3^{(2)}$. In a mixed term one subsystem's magnetic quantum number increases and the other's decreases, so its commutator with total $J_3$ is zero. Hence $\boxed{[{\mathbf{J}}^2,J_3]=0}$.

Use the standard identities $[J_3,J_{\pm }]=\pm J_{\pm }$ and $J_{\pm }|j,m⟩=\sqrt{(j\mp m)(j\pm m+1)}|j,m\pm 1⟩$. The PDF has the commutator in the reversed order with unchanged sign and prints $m+1$ also for lowering; these are sign/label slips. The standard identities are consistent with the meaning of raising and lowering and are used here.

Put $s=j_1+j_2$. The unique product state of magnetic quantum number $-s$ isFor $j_1,j_2>0$, define the orthonormal product states $|a⟩=|j_1,-j_1+1⟩|j_2,-j_2⟩$ and $|b⟩=|j_1,-j_1⟩|j_2,-j_2+1⟩$. Raising the preceding state givesThe normalized orthogonal combination is annihilated by $J_{-}$, because its two lowering amplitudes cancel. It is consequently the lowest-weight state with $J=s-1$:up to an overall phase. For $j_1=3,j_2=1$, the outcome $m_1=-3,m_2=0$ is the $|b⟩$ component. The Born rule therefore gives