= Solution
Writing matrices in $2\times2$ blocks, the defining relation gives
$$
L=\left\{\begin{pmatrix}A&B\\C&-A^T\end{pmatrix}:B^T=-B,\ C^T=-C\right\}.
$$
The <trace> is automatically zero. The relation is linear, and if $as=-sa^T$, $bs=-sb^T$, then
$$
[a,b]s=s(a^Tb^T-b^Ta^T)=-s[a,b]^T.
$$
Thus $L$ is closed under the <commutator> and is a <Lie subalgebra> of $\mathfrak{sl}_4(F)$.
Its diagonal subalgebra is
$$
H=\{D(a,b)=\operatorname{diag}(a,b,-a,-b):a,b\in F\}.
$$
Put $\varepsilon_1(D(a,b))=a$, $\varepsilon_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition is
$$
\boxed{\begin{aligned}
L={}&H\oplus F(E_{12}-E_{43})\oplus F(E_{21}-E_{34})\\
&\oplus F(E_{14}-E_{23})\oplus F(E_{32}-E_{41}).
\end{aligned}}
$$
The four lines have respective characters $\varepsilon_1-\varepsilon_2$, $-\varepsilon_1+\varepsilon_2$, $\varepsilon_1+\varepsilon_2$, $-\varepsilon_1-\varepsilon_2$. They are distinct and nonzero because $\operatorname{char}F\ne2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the <dimension> obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.
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