= Solution
Under the two-block <subgroup> $H_2$, the defining <vector space> splits into its first and last two coordinates. Therefore
$$
\boxed{\begin{aligned}
\mathbf4\downarrow_{H_2}&=(\mathbf2,\mathbf1)\oplus(\mathbf1,\mathbf2),\\
\mathbf6\downarrow_{H_2}&=(\mathbf1,\mathbf1)^{\oplus2}\oplus(\mathbf2,\mathbf2),\\
\mathbf{15}\downarrow_{H_2}&=(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3)
\oplus(\mathbf2,\mathbf2)^{\oplus2}\oplus(\mathbf1,\mathbf1).
\end{aligned}}
$$
For the <exterior square>, the <exterior square of a direct sum> identity gives two within-block <determinant representations>, each trivial for $SU(2)$, and the cross-block <tensor product> $(\mathbf2,\mathbf2)$. For the <Adjoint representation>, use $\operatorname{End}(\mathbb C^4)-\mathbf1$: each diagonal two-by-two block contributes a scalar and a traceless triplet, while each off-diagonal block gives $(\mathbf2,\mathbf2)$ because the <SU(2)> doublet is self-dual. Removing the overall <trace> leaves one scalar. The <dimension> counts are $2+2=4$, $1+1+4=6$, and $3+3+4+4+1=15$.
Under $H_3$, the first three coordinates carry the defining <SU(3)> <representation> and the fourth is fixed. Hence
$$
\boxed{\begin{aligned}
\mathbf4\downarrow_{H_3}&=\mathbf3\oplus\mathbf1,\\
\mathbf6\downarrow_{H_3}&=\overline{\mathbf3}\oplus\mathbf3,\\
\mathbf{15}\downarrow_{H_3}&=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1.
\end{aligned}}
$$
Indeed $\wedge^2(\mathbf3\oplus\mathbf1)=\wedge^2\mathbf3\oplus\mathbf3=\overline{\mathbf3}\oplus\mathbf3$. The upper three-by-three block of a traceless <matrix> supplies the <Adjoint representation> $\mathbf8$ and one scalar, while the upper-right and lower-left blocks give $\mathbf3$ and $\overline{\mathbf3}$. This establishes the <branching of a representation> in each case without a dimension-only guess.
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