= Solution
Under the subgroup in part (b), $(v,s)\in\mathbb R^4\oplus\mathbb R$ transforms as $(Rv,s)$, hence
$$
\mathbf5\downarrow SO(4)=\mathbf4\oplus\mathbf1.
$$
An element of $\mathfrak{so}_5$ has the unique block form
$$
X=\begin{pmatrix}A&v\\-v^T&0\end{pmatrix},
\qquad A\in\mathfrak{so}_4,quad v\in\mathbb R^4.
$$
Conjugation by $\iota(R)$ sends $(A,v)$ to $(RAR^{-1},Rv)$. Thus
$$
\mathbf{10}\downarrow SO(4)=\mathbf6_{\mathrm{ad}}\oplus\mathbf4,
$$
and the dimensions check as $5=4+1$ and $10=6+4$. These are the <SO5 to SO4 branching> rules.
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