= Solution
Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\ldots,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in\{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.
For $j=1/2$, abbreviate the product state $|1,q\rangle|1/2,m\rangle$ by $|q,\uparrow\rangle$ or $|q,\downarrow\rangle$. The highest-weight state is $|3/2,3/2\rangle=|1,\uparrow\rangle$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:
$$
\begin{aligned}
|\tfrac32,\tfrac32\rangle&=|1,\uparrow\rangle,\\
|\tfrac32,\tfrac12\rangle&=\sqrt{\tfrac23}|0,\uparrow\rangle+\sqrt{\tfrac13}|1,\downarrow\rangle,\\
|\tfrac32,-\tfrac12\rangle&=\sqrt{\tfrac23}|0,\downarrow\rangle+\sqrt{\tfrac13}|-1,\uparrow\rangle,\\
|\tfrac32,-\tfrac32\rangle&=|-1,\downarrow\rangle.
\end{aligned}
$$
The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it gives
$$
\begin{aligned}
|\tfrac12,\tfrac12\rangle&=\sqrt{\tfrac13}|0,\uparrow\rangle-\sqrt{\tfrac23}|1,\downarrow\rangle,\\
|\tfrac12,-\tfrac12\rangle&=\sqrt{\tfrac23}|-1,\uparrow\rangle-\sqrt{\tfrac13}|0,\downarrow\rangle.
\end{aligned}
$$
These six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding <Clebsch-Gordan coefficients>.
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