= Solution
For the <branching from B3 to A3>, apply the supplied <weight projection matrix> to column <vectors> of <Dynkin labels>. It sends
$$
\boxed{(a_1,a_2,a_3)^T\longmapsto(a_2,a_1,a_2+a_3)^T.}
$$
It is consistent with choosing the $A_3$ <simple roots> in the order $\beta_1=e_2-e_3$, $\beta_2=e_1-e_2$, $\beta_3=e_2+e_3$. Their <Dynkin diagram> is the chain $\beta_1-\beta_2-\beta_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.
The seven <weights> in <4/b> project as
$$
\begin{array}{c|c}
B_3&A_3\\\hline
(1,0,0)&(0,1,0)\\
(-1,1,0)&(1,-1,1)\\
(0,-1,2)&(-1,0,1)\\
(0,0,0)&(0,0,0)\\
(0,1,-2)&(1,0,-1)\\
(1,-1,0)&(-1,1,-1)\\
(-1,0,0)&(0,-1,0)
\end{array}.
$$
The six nonzero <weights> form the <weight system> of the <SU(4)> <two-index antisymmetric representation> of <highest weight> $(0,1,0)$. The zero <weight> is the remaining <singlet representation>. Consequently
$$
\boxed{\mathbf7\downarrow_{A_3}=\mathbf6\oplus\mathbf1.}
$$
The eight spinor <weights> project as
$$
\begin{array}{c|c}
B_3&A_3\\\hline
(0,0,1)&(0,0,1)\\
(0,1,-1)&(1,0,0)\\
(1,-1,1)&(-1,1,0)\\
(1,0,-1)&(0,1,-1)\\
(-1,0,1)&(0,-1,1)\\
(-1,1,-1)&(1,-1,0)\\
(0,-1,1)&(-1,0,0)\\
(0,0,-1)&(0,0,-1)
\end{array}.
$$
The four <weights> $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining <fundamental representation> $\mathbf4$. The other four are their negatives, forming the <dual representation> $\overline{\mathbf4}$ of <highest weight> $(0,0,1)$. Hence
$$
\boxed{\mathbf8\downarrow_{A_3}=\mathbf4\oplus\overline{\mathbf4}.}
$$
For the <Young tableaux> and <tensors>, let $U=\mathbb C^4$ be the defining <SU(4)> <representation>. The <singlet representation> is $\Lambda^0U$, with the empty <Young diagram> and a scalar. The other three <Young diagrams> are columns of one, two and three boxes, representing $U$, $\Lambda^2U$ and $\Lambda^3U$:
$$
\begin{array}{c|c|c|c}
\text{highest weight}&\text{representation}&\text{tensor}&\text{dimension}\\\hline
(0,0,0)&\mathbf1&S&1\\
(1,0,0)&\mathbf4&v^i&4\\
(0,1,0)&\mathbf6&T^{[ij]}&\binom42=6\\
(0,0,1)&\overline{\mathbf4}&T^{[ijk]}&\binom43=4
\end{array}.
$$
The square brackets indicate <tensor index antisymmetrization>, with indices from 1 to 4. The invariant <Levi-Civita symbol> identifies the three-index <antisymmetric tensor> with a covector, $w_\ell=\epsilon_{\ell ijk}T^{ijk}/3!$, explaining $\Lambda^3U\cong U^*$.
\Image[/past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2018-su4-young-tableaux.png]
{title=Young tableaux for the SU4 representations in the two branchings}
{description=The empty <Young diagram> gives the <singlet representation>; columns of one, two and three boxes give $\mathbf4$, $\mathbf6$ and $\overline{\mathbf4}$. Increasing entries down each column display a standard <Young tableau>. The column determines complete <tensor index antisymmetrization>.}
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