= 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 $J_\pm^{(i)}=J_1^{(i)}\pm iJ_2^{(i)}$. Expanding the square gives
$$
\boxed{\mathbf J^2=(\mathbf J^{(1)})^2+(\mathbf J^{(2)})^2+2J_3^{(1)}J_3^{(2)}+J_+^{(1)}J_-^{(2)}+J_-^{(1)}J_+^{(2)}.}
$$
The 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 $J_\pm|j,m\rangle=\sqrt{(j\mp m)(j\pm m+1)}|j,m\pm1\rangle$. 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 $-s$ is
$$
\boxed{|s,-s\rangle=|j_1,-j_1\rangle|j_2,-j_2\rangle.}
$$
For $j_1,j_2>0$, define the orthonormal product states $|a\rangle=|j_1,-j_1+1\rangle|j_2,-j_2\rangle$ and $|b\rangle=|j_1,-j_1\rangle|j_2,-j_2+1\rangle$. Raising the preceding state gives
$$
|s,-s+1\rangle=\sqrt{\frac{j_1}s}|a\rangle+\sqrt{\frac{j_2}s}|b\rangle.
$$
The normalized orthogonal combination is annihilated by $J_-$, because its two lowering amplitudes cancel. It is consequently the lowest-weight state with $J=s-1$:
$$
\boxed{|s-1,-s+1\rangle=\sqrt{\frac{j_2}s}|a\rangle-\sqrt{\frac{j_1}s}|b\rangle,}
$$
up to an overall phase. For $j_1=3,j_2=1$, the outcome $m_1=-3,m_2=0$ is the $|b\rangle$ component. The <Born rule> therefore gives
$$
\boxed{\mathbb P(m_1=-3,m_2=0\mid|3,-3\rangle)=\frac34.}
$$
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