= Solution
For the block computation only, place the $f$-vectors in the order $f_1,\ldots,f_m$ after the $e$-vectors. This temporary reversal of the second block changes no transformations. The form matrix becomes $J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}$.
Let $T$ be upper unitriangular of size $m$, and let $S$ be symmetric. The matrices
$$
d(T)=\begin{pmatrix}T&0\\0&T^{-T}\end{pmatrix},\qquad
n(S)=\begin{pmatrix}I&S\\0&I\end{pmatrix}
$$
satisfy $gJg^T=J$ by direct block multiplication. The generator $x_{ij}(\lambda)$ is $d(I+\lambda E_{ij})$, because the inverse transpose adds $-\lambda f_i$ to $f_j$. The generators $y_{ij}(\lambda)$ and $z_i(\lambda)$ are respectively $n(\lambda(E_{ij}+E_{ji}))$ and $n(\lambda E_{ii})$. This proves all of them preserve the form, in every characteristic.
The $x$-generators generate every upper unitriangular $T$, by elimination of off-diagonal entries. The $y,z$ generators add all elementary symmetric entries, so they generate every $n(S)$. Moreover
$$
d(T)n(S)d(T)^{-1}=n(TST^T),
$$
which keeps symmetry. Therefore the generated group is exactly
$$
\{n(S)d(T):S=S^T,\ T\text{ upper unitriangular}\}.
$$
The two factors are uniquely determined by its diagonal and off-diagonal blocks. Their counts are $q^{m(m+1)/2}$ and $q^{m(m-1)/2}$, giving
$$
\boxed{|U_0|=q^{m^2}.}
$$
It is a $p$-group, and this equals the full $p$-part found in part (a); hence $U_0$ is a Sylow $p$-subgroup. In the original reversed-$f$ ordering, these matrices are upper unitriangular throughout, exactly as the prescribed generators suggest. This is consistent with the <finite symplectic group order>. No factor of two was divided out, so characteristic two is included.
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