I'm trying to understand model in a propositional logic. Assume I have a set of formulas $F$, and a formula $\psi$. What exactly does $F \models \psi$ mean here? Does it only require that I have a single formula in $F$, such that it is model of $\psi$? Then, if I have $\psi \in F$, it must hold that $F \models \psi$, right?
Also, assume I have a formula $\phi = q_1 \wedge \cdots \wedge q_n \rightarrow p$. Can I say $\phi \models p$, if we have that $q_1,\ldots,q_n = TRUE$, as $p$ is implied by $q_1,\ldots,q_n$?
$F\models \psi$ means that any interpretation where all the formulas in $F$ are satisfied also satisfies $\psi.$ If $\psi\in F$ then certainly $F\models \psi$, since any interpretation where all the formulas in $F$ are satisfied satisfies $\psi$ (since $\psi$ is in $F$). However, it is possible for $F\models \psi$ but $\psi\notin F.$ A silly example of this is if $F=\{p_1\land p_2\},$ then we have $F\models p_1$ but $p_1\notin F.$ More generally, a set of formulas will usually have more consequences than just the formulas themselves.
No, you can't say $\phi\models p.$ It's true that if $q_1,\ldots q_n$ are all true then $p$ is true, but the assumption $\phi$ does not imply that $q_1,\ldots, q_n$ are all true. In fact, an interpretation where one of the $q_i$ is false and $p$ is false will satisfy $\phi$ but not $p$, so assuming the $q_i$ and $p$ are atomic letters that can be independently set true or false, we have $\phi\not\models p.$ What is true is that $\phi,q_1,\ldots, q_n\models p$ or equivalently that $\phi,q_1\land q_2\ldots\land q_n\models p.$ More generally, if $\phi:=A\to B$ (or even just $\phi\models A\to B$) then we have $\phi,A\models B.$