3 solved out of 5 shown

Is it true that the number of graphs on $n$ vertices which do not contain $G$ is
\[\leq 2^{(1+o(1))\mathrm{ex}(n;G)}?\]

If $G$ is not bipartite the answer is yes, proved by Erdős, Frankl, and Rödl [ErFrRo86]. The answer is no for $G=C_6$, the cycle on 6 vertices. Morris and Saxton [MoSa16] have proved there are at least
\[2^{(1+c)\mathrm{ex}(n;C_6)}\]
such graphs for infinitely many $n$, for some constant $c>0$. It is still possible (and conjectured by Morris and Saxton) that the weaker bound of
\[2^{O(\mathrm{ex}(n;G))}\]
holds for all $G$.

If $\mathcal{F}$ is a finite set of finite graphs then $\mathrm{ex}(n;\mathcal{F})$ is the maximum number of edges a graph on $n$ vertices can have without containing any subgraphs from $\mathcal{F}$. Note that it is trivial that $\mathrm{ex}(n;\mathcal{F})\leq \mathrm{ex}(n;G)$ for every $G\in\mathcal{F}$.

Is it true that, for every $\mathcal{F}$, there exists some constant $C_{\mathcal{F}}>0$ and $G\in\mathcal{F}$ such that for all large $n$ \[\mathrm{ex}(n;G)\leq C_{\mathcal{F}}\mathrm{ex}(n;\mathcal{F})?\]

This is trivially true if $\mathcal{F}$ does not contain any bipartite graphs, since by the Erdős-Stone theorem if $H\in\mathcal{F}$ has minimal chromatic number $r\geq 2$ then
\[\mathrm{ex}(n;H)=\mathrm{ex}(n;\mathcal{F})=\left(\frac{r-2}{r-1}+o(1)\right)\binom{n}{2}.\]
Erdős and Simonovits observe that this is false for infinite families $\mathcal{F}$, e.g. the family of all cycles.