We can obtain a linear map $V \to W^*$ by sending $x \in V$ to the functional $y \mapsto F(x, y)$ on $W$. That $F$ is non-degenerate implies that this map is injective, so $\dim V \leq \dim W^*$. Since $W$ is finite-dimensional, $W^*$ has the same dimension as $W$ and hence $\dim V \leq \dim W$. Using the analogous map $W \to V^*$, we get the reverse inequality.
I try to answer (2). You can try to think about (1) afterwards.
Define $U := \{u \in V: J(u,v) = v \text{ for any } v \in V \}$. Since $J$ is non-degenerate, then $U = \{0\}$. So, if $V$ is not trivial, there exists an element $e_1 \in V \setminus \{0\}$ such that $J(e_1,f_1) \neq 0$ for some $f_1 \in V$. Up to rescaling, you can assume $J(e_1,f_1) = 1$. Call $W := \text{Span}(e_1,f_1)$ and define
$$W^J := \{u \in V: J(u,w) = 0 \text{ for any }w \in W\}.$$ Let us take a look at $W \cap W^J$. If $v \in W \cap W^J$, then $v = ae_1+bf_1$ for some $a,b$, and $J(v,e_1) = 0 = J(v,f_1)$. But then $J(v,e_1)=J(ae_1+bf_1,e_1) = -b=0$ and similarly $a=0$. So $W \cap W^J = \{0\}$. Let now $v$ be any vector in $V$. If $J(v,e_1) = -a$ and $J(v,f_1) = b$, then you can write
$$v = be_1+af_1+v-be_1-af_1.$$
You have that $be_1+af_1 \in W$ and
\begin{align}
J(v-be_1-af_1,e_1) & = J(v,e_1)+a = a-a=0\\
J(v-be_1-af_1,f_1) & = J(v,f_1)-b = b-b = 0.
\end{align}
This tells you that any vector $v \in V$ can be written as a sum of a vector in $W$, that is $be_1+af_1$, and a vector in $W^J$, namely $v-be_1-af_1$. Consequently $V = W \oplus W^J$. If $W^J = \{0\}$, then you are done and $J$ can be written in the basis $\{e_1,f_1\}$ as
$$
\left(
\begin{matrix}
J(e_1,e_1) & J(e_1,f_1) \\
J(f_1,e_1) & J(f_1,f_1)
\end{matrix}
\right) =
\left(
\begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix}
\right).
$$
Otherwise, choose $e_2 \neq 0$ in $W^J$ and repeat the process getting $f_2$ such that $J(e_2,f_2)=1$. Going on you will find a basis $\{e_1,e_2,\dots,e_n,f_1,f_2,\dots,f_n\}$ of $V$ such that
$$J(e_i,e_j) = 0, \quad J(e_i,f_k) = \delta_{ik}, \quad J(f_i,f_j) = 0,$$ where $\delta_{ij}$ denotes the Kronecker delta. The process ends after $n$ steps, as $\dim V < \infty$. Notice that the presence of $J$ forces $\dim V = 2n$, i.e. the dimension of $V$ is even.
FYI: a non-degenerate skew-symmetric bilinear form $J$ like this is generally called symplectic form on $V$, and $(V,J)$ is then called symplectic space.
Best Answer
Note that this bilinear form is not degenerate. Indeed you take $B=^tA $ and obtain $\phi : \mathbb{M_{n}(R)}*\mathbb{M_{n}(R)} \rightarrow \mathbb{R}$ $$(A,^tA) \rightarrow trace(A^tA)\ge0$$ To find the orthogonal of symmetric matrix note that if $A $ is a anti-symmetric matrix then $$(A,B) =tr(AB)=tr(^t(BA))=tr(^tA^tB)=tr(-AB)=-(A,B)$$ from here $$(A,B)=0$$ Note that the bilinear form is not degenerate we obtain that the dimension of orthogonal of symmetric matrix is given by $n^2$- the dimension of symmetric matrix. You obtain that $$dim \mathbb{S_{n}(R)}^{\perp}=n^2-dim \mathbb{S_{n}(R)}=dim\mathbb{A_{n}(R)}$$ where $\mathbb{A_{n}(R)}$ are anti-symmetric matrix.