First I want to review some concept from quadratic form.
Let $V$ be quadratic space over finite field $F$ and $char(F)\neq 2$ with quadratic form $q$. For exmaple


$q:V\rightarrow V$ and $|F|=q$ and $\dim V=n$.


We may assume the vectors of $V$ are represented as column vectors $v=(a\_1,b\_1,…)\in V$, and that $q$ takes one of the three forms


1.$q(v)={a\_1}^2-{b\_1}^2+…+{a\_k}^2-{b\_k}^2$


2.$q(v)={a\_1}^2-{b\_1}^2+…+{a\_k}^2-{db\_k}^2$


3.$q(v)={a\_1}^2-{b\_1}^2+…+{a\_k}^2-{b\_k}^2-{a\_{k+1}}^2$


or if we want to have matrix representation we have


1.$q=diag(1,-1,1,-1,…,1,-1)$


2.$q=diag(1,-1,1,-1,…,1,-1,1)$


3.$q=diag(1,-1,1,-1,…,1,-1,1,-d)$


Finally for subspace $W$ we have $W^{\perp}=\{v\in V\mid b(v,w)=0 ,\forall w \in W \}$ where $b(x,y)=(\frac{1}{2})[q(x+y)-q(x)-q(y)]$.


The problem I am interested in :


Compute the number of $k$-dimensional subspaces $W$ such that $W\subseteq W^{\perp}$; this subspace is called **totally isotropic** subspace (T,I). For $k=1$, it is prominent theorem. That comes back to find solution of these equations


1.$q(v)={x\_1}^2-{y\_1}^2+…+{x\_k}^2-{y\_k}^2=0$


2.$q(v)={x\_1}^2-{y\_1}^2+…+{x\_k}^2-{dy\_k}^2=0$


3.$q(v)={x\_1}^2-{y\_1}^2+…+{x\_k}^2-{y\_k}^2-{x\_{k+1}}^2=0$
So the number of 1-dimensional (T,I) is


1.$\frac{q^{2k-1}+q^{k}-q^{k-1}-1}{q-1}$


2.$\frac{q^{2k-1}-q^{k}+q^{k-1}-1}{q-1}$


3.$\frac{q^{2k}-1}{q-1}$


Now what can I do for $k\geq 2$.



Number of Totally Isotropic Subspaces
Let $q$ denote the order of the finite field, which assume has odd characteristic. If $Q$ is a non-degenerate quadratic form in $n$ variables, for $n$ odd (this is your case 3), we have that the number of totally isotropic subspaces of dimension $k$ (for $k \leq \frac{n-1}{2}$, otherwise the number is $0$) is 
$$ \frac{\prod\_{i=0}^{k-1} (q^{2n-2i-1 } -1)}{\prod\_{i=1}^{k} (q^i -1)}.$$


If the number of variables, $n$, is even then the answer depends on the signature of the quadratic form. Let $\epsilon = \{+1,-1\}$ denote this signature. The number of totally isotropic subspaces of dimension $k$ is then


$$ \frac{(q^{n-k} - \epsilon q^{n/2-k} + \epsilon q^{n/2} -1) \prod\_{i=1}^{k-1} (q^{n-2i} -1 ) }{\prod\_{i=1}^{k} (q^i -1) }.$$


Again, we assume that $k\leq n/2-1$ if $\epsilon =-1$ and $k \leq n/2$ if $\epsilon = +1$ (otherwise the number is $0$). Note that $\epsilon=+1$ corresponds to your case $1$ and $\epsilon=-1$ corresponds to your case $2$.


A short proof of these identities can be found in:


V. Pless, [The number of isotropic subspaces in a finite geometry](http://www.ams.org/mathscinet-getitem?mr=199785). Atti Accad. Naz. Lincei Rend. CI. Sci. Fic. Mat. Natur. (8) 39 1965 418-421.


although I believe these results were known previously.


