Sam is correct of course about $q$-hook formula. Below is a short self-contained proof not relying on such advanced combinatorics.
Denote $h_1>\ldots>h_k$ the set of hook lengths of the first column of diagram $\lambda$. Then the multiset of hooks is $\cup_{i=1}^k \{1,2,\ldots,h_i\}\setminus \{h_i-h_j:i<j\}$ and $n=\sum_i h_i-\frac{k(k-1)}2$.
Recall that $F(m)=P_m(\alpha,\beta)=\prod_{d|m,d>1}\Phi_d(\alpha,\beta)=\prod_d (\Phi_d(\alpha,\beta))^{\eta_d(m)}$, where
$\alpha,\beta=(1\pm \sqrt{5})/2$;
$P_n(x,y)=x^{n-1}+x^{n-2}y+\ldots+y^{n-1}$;
$\Phi_d$ are homogeneous cyclotomic polynomials;
$\eta_d(m)=\chi_{\mathbb{Z}}(m/d)$ (i.e., it equals to 1 if $d$ divides $m$, and to 0 otherwise).
Therefore it suffices to prove that for any fixed $d>1$ we have
$$
\sum_{m=1}^n \eta_d(m)+\sum_{i<j}\eta_d(h_i-h_j)-\sum_{i=1}^k\sum_{j=1}^{h_i}\eta_d(j)\geqslant 0.\quad (\ast)
$$
$(\ast)$ rewrites as
$$
[n/d]+|i<j:h_i\equiv h_j \pmod d|-\sum_{i=1}^k [h_i/d]\geqslant 0.\quad (\bullet)
$$
LHS of $(\bullet)$ does not change if we reduce all $h_i$'s modulo $d$ (and accordingly change $n=\sum_i h_i-\frac{k(k-1)}2$, of course), so we may suppose that $0\leqslant h_i\leqslant d-1$ for all $i$. For $a=0,1,\dots, d-1$ denote $t_a=|i:h_i=a|$. Then $(\bullet)$ rewrites as
$$
\left[\frac{-\binom{\sum_{i=0}^{d-1} t_i}2+\sum_{i=0}^{d-1} it_i}d\right]+
\sum_{i=0}^{d-1} \binom{t_i}2\geqslant 0. \quad (\star)
$$
It remains to observe that LHS of $(\star)$ equals to
$$
\left[\frac1d\sum_{i<j}\binom{t_i-t_j}2 \right].
$$
Proposition. A partition $\lambda \in \mathcal{syP}_0$ iff it is empty, or both of the following hold:
- $\lambda'_1 - \lambda_1 = 1$,
- the partition $\mu$ obtained by removing first row and column of $\lambda$ is also in $\mathcal{syP}_0$.
Informally, all partitions in $\mathcal{syP}_0$ (and only those) are obtained by starting from the empty partition, and repeatedly adding "L-shapes" to the bottom left corner, each L-shape one cell wider than it is taller, given that each intermediate step yields a proper (column-monotonic) partition. For example, $\lambda = 43221$ looks like this:
A
AB
ABBB
AAAAA
and $\lambda = 3333$ looks like this
ABCC
ABBB
AAAA
To see this, reverse the sequence $\lambda_0 = \lambda, \lambda_1 = \mu, \ldots, \lambda_k = \varnothing$, where each subsequent partition is obtained by "shedding" the first row and column of the one before it. As explained below in the proof, if at any point $\lambda'_{i, 1} - \lambda_{i, 1} \neq 1$, then we present a cell with $c_{sp}(\square) = 0$, otherwise the claim is established by induction.
Proof. Observe that symplectic content of cells of $\mu$ is carried over to corresponding cells of $\lambda$, as such it has to be/stays non-zero.
If $\lambda'_1 - \lambda_1 > 1$, then $c_{sp}(i, j) = 0$ for $(i, j) = (\lambda_1 + 2, 1)$. Indeed $i > j$ and $\lambda_i = 1$, and $c_{sp}(i, j) = 1 + \lambda_1 - (\lambda_1 + 2) - 1 + 2 = 0$.
If $\lambda'_1 - \lambda_1 < 1$, then similarly $c_{sp}(1, \lambda'_1) = 1+\lambda'-\lambda'-1=0$.
Finally, if $\lambda'_1 - \lambda_1 = 1$, then for any $x \in \{1, \ldots, \lambda_1\}$ we have $c_{sp}(1, x) = 1 + x - \lambda'_1 - \lambda'_x \leq 1 + \lambda_1 - \lambda'_1 - \lambda'_x = -\lambda'_x < 0$. Similarly one obtains that $c_{sp}(x + 1, 1)=\lambda_{x+1}+\lambda_1-1-(x+1)+2=\lambda_{x+1}+\lambda_1-x\geq \lambda_{x+1} > 0$. $\square$
The bijection to partitions into even distinct parts is now obvious: use the parts as descending even sizes (hook-lengths $h(i,i)$ of the diagonals) of L-shapes, which answers Q1.
Q2 is now also easy to answer in the positive. Both $h(\square)$ and $c_{sp}(\square)$ are preserved after adding/removing the first row and column. Direct substitution yields $h(1, x) = -c_{sp}(1, x)$, $h(x + 1, 1) = c_{sp}(x + 1, 1)$ iff we are allowed to substitute $\lambda'_1 = \lambda_1 + 1$.
Best Answer
A hook-content formula, using the contents $c_{sp}(i,j)$ and $c_O(i,j)$, for the dimensions of the irreducible polynomial representations of the symplectic and orthogonal groups, goes back to Ron King. I believe the relevant paper is https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0008414X00053086, but I did not check for sure.