= Solution
The <Clifford algebra> relation implies
$$
[C(a)C(b),C(v)]=2(b,v)C(a)-2(a,v)C(b).
$$
Indeed move $C(v)$ past $C(b)$ and then past $C(a)$ using the two anticommutator identities; the three-generator products cancel.
Since all $C(w)$ are <self-adjoint>,
$$
\pi(T)^*+\pi(T)=\frac14\sum_i\bigl(C(e_i)C(Te_i)+C(Te_i)C(e_i)\bigr)
=\frac12\sum_i(e_i,Te_i)I=0.
$$
Each scalar vanishes because $T$ is <skew-adjoint>. This proves $\boxed{\pi(T)^*=-\pi(T)}$.
For the requested <commutator>, the first identity gives
$$
[\pi(T),C(v)]=\frac12\sum_i\bigl((e_i,v)C(Te_i)-(Te_i,v)C(e_i)\bigr).
$$
The first sum is $C(Tv)$ by the <orthonormal basis> expansion of $v$. For the second, the <skew-adjoint> property gives $\sum_i(Te_i,v)e_i=-Tv$. By real linearity of $C$, the difference is therefore
$$
\boxed{[\pi(T),C(v)]=\tfrac12(C(Tv)-C(-Tv))=C(Tv).}
$$
This calculation explains the factor $1/4$: it makes infinitesimal rotations of $V$ act correctly on its Clifford generators.
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