= Solution
Replace the <determinant> condition by preservation of a nondegenerate <alternating bilinear form> $B$ on $F^{2m}$. In a <symplectic basis> the <symplectic group over a finite field> is
$$
Sp_{2m}(q)=\{g\in GL_{2m}(q):g^{\mathsf T}Jg=J\},
\qquad J=\begin{pmatrix}0&I_m\\-I_m&0\end{pmatrix}.
$$
To count <symplectic bases>, choose the first <vector> $e$ in $q^{2m}-1$ ways, then choose $f$ with $B(e,f)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate; its <symplectic orthogonal complement> has <dimension> $2m-2$ and supplies the rest of the <basis>. The resulting recurrence gives the <finite symplectic group order>
$$
\boxed{|Sp_{2m}(q)|=q^{m^2}\prod_{i=1}^m(q^{2i}-1).}
$$
The elementary replacements are <symplectic transvections>
$$
T_{v,c}(x)=x+cB(x,v)v,\qquad v\ne0,\ c\in F.
$$
Expanding $B(Tx,Ty)$ shows that the two cross terms cancel and the last term vanishes because $B(v,v)=0$. Thus these maps preserve $B$; their inverses are $T_{v,-c}$.
We need both generation and primitivity to use the argument from part (i). For nonzero <vectors> $v,w$ with $B(v,w)\ne0$, the <symplectic transvection> with direction $w-v$ and parameter $1/B(v,w)$ sends $v$ to $w$. When $B(v,w)=0$, choose $z$ nonorthogonal to both and use two such steps. Such $z$ exists because two proper <hyperplanes> cannot cover a <vector space>: if the two linear forms are independent prescribe both values to be one, and if dependent prescribe either a nonzero value. Hence <symplectic transvections> act transitively on nonzero <vectors>.
They generate the whole <symplectic group over a finite field>. Indeed first use them to make a given symplectic transformation fix $e_1$. Its image $f'$ of $f_1$ still satisfies $B(e_1,f')=1$. A <symplectic transvection> with direction $f_1-f'$ fixes $e_1$ and sends $f'$ to $f_1$ if the pairing is nonzero. If that pairing is zero, pass through $f'+ce_1$ for any $c\ne0$: it has nonzero pairing with both partners, and both required directions lie in $e_1^\perp$. We have now fixed the first <hyperbolic pair>, and induction applies in its nondegenerate orthogonal complement. The final two-dimensional case is exactly the elementary generation of $SL_2(q)=Sp_2(q)$ proved in part (i).
For a fixed <projective point> $\langle e\rangle$, the <symplectic transvections> $T_{e,c}$ form an abelian <subgroup> normal in its <stabilizer subgroup>; conjugation scales $c$ by the square of the scalar multiplying $e$. Their conjugates supply all <symplectic transvections>. A central element commutes with every <symplectic transvection>, so it preserves every direction line and must be scalar. Conversely scalars commute with the whole group. The kernel of the projective action consists of scalar <matrices>, and preservation of $B$ restricts the scalar to $\lambda^2=1$. Thus the <center of a group> has size $\gcd(2,q-1)$ and the <projective symplectic group over a finite field> acts faithfully.
For $m\ge2$ the <stabilizer subgroup> has precisely three orbits: its fixed point, the other points in $e^\perp$, and the points outside $e^\perp$. Transitivity on each follows by extending, respectively, an independent isotropic pair or a <hyperbolic pair> to a <symplectic basis>. Write
$$
N=\frac{q^{2m}-1}{q-1},\qquad
h=\frac{q^{2m-1}-1}{q-1}.
$$
A nontrivial proper block through the fixed point would be a union of these <stabilizer subgroup> orbits, hence would have size either $h$ or $q^{2m-1}+1$. A block size divides $N$. But $N-qh=1$, so $h>1$ cannot divide $N$, and $N/2<q^{2m-1}+1<N$, excluding the other size. This proves primitivity. For $m=1$ the projective action is the <two-transitive> action of $PSL_2(q)$.
Here is a direct check of <perfectness of finite symplectic groups>, including the small fields where the diagonal <group commutator> needs replacement. If $q>3$, every <symplectic transvection> lies in an embedded $SL_2(q)$ on a hyperbolic plane containing its direction, and that <subgroup> is perfect by part (i). Since the <symplectic transvections> generate, the whole <symplectic group over a finite field> is perfect.
For $q=3$ and $m\ge2$, all $T_{v,1}$ are conjugate, so their common image $z$ generates the <abelianization> and satisfies $3z=0$. Choose independent perpendicular <vectors> $e,f$. The four commuting <symplectic transvections> with directions $e,f,e+f,e-f$ and parameter one multiply to the identity: the sum of their rank-one <matrices> is zero in characteristic 3. Thus also $4z=0$, giving $z=0$.
For $q=2$ and $m\ge3$, use the seven nonzero <vectors> of a three-dimensional <isotropic subspace of a symplectic vector space>. Their <symplectic transvections> commute, and their product is the identity. To see this explicitly, write a <vector> in three isotropic <basis> coordinates: in the sum of the outer products $vv^{\mathsf T}$ each diagonal coefficient occurs four times and each off-diagonal coefficient twice, all zero in characteristic 2. The common <abelianization> image $z$ therefore satisfies $2z=7z=0$. Again the group is perfect.
The simplicity argument from part (i) now applies to the projective <symplectic group over a finite field>. The remaining exceptional group besides $Sp_2(2)$ and $PSp_2(3)$ is $Sp_4(2)$. It is not simple: it is isomorphic to $S_6$. One can verify this identification without assuming it. On $F_2^4$ set
$$
Q_0(x_1,x_2,y_1,y_2)=x_1y_1+x_2y_2,\qquad
Q_a(v)=Q_0(v)+B(a,v).
$$
These are all 16 <quadratic forms> with polar form $B$; the <Arf invariant of a quadratic form> gives $\operatorname{Arf}(Q_a)=Q_0(a)$. There are six with invariant one. The <symplectic group over a finite field> permutes these six forms. Differences of their parameter <vectors> span $F_2^4$: writing $e_i,f_i$ for the <symplectic basis>, the six parameters are
$$
e_1+f_1,\ e_1+f_1+e_2,\ e_1+f_1+f_2,\
e_2+f_2,\ e_2+f_2+e_1,\ e_2+f_2+f_1.
$$
Their pairwise differences include each of $e_1,e_2,f_1,f_2$. An element fixing all six forms therefore fixes all linear forms $B(a-b,-)$, hence is the identity. This gives a faithful action on six objects; the group order is 720, so its image is all of $S_6$. Consequently
$$
\boxed{|PSp_{2m}(q)|=\frac{q^{m^2}\prod_{i=1}^m(q^{2i}-1)}{\gcd(2,q-1)},\qquad
PSp_{2m}(q)\text{ simple except }(m,q)=(1,2),(1,3),(2,2).}
$$
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