Let $F \subset B$ be a proper filter. Prove that $F$ is maximal if and only if for all $p \in B$ with the property that $p \wedge q \ne 0$, for all $q \in F$, the $p$ is also in $F$. I've used a characterisation of maximal filter which states that a filter $F$ is maximal if and only if either $p$ or $p^c$ are in $F$, for all $p \in B$
What I've proved is the following:
Let F be a maximal filter. Suppose there is a $p \in B$, such that $p \wedge q \ne 0$, for all $q \in B$, but $p \notin F$. Because the $F$ is maximal and $p \notin F$ it must be $p^c \in F$. But now $0 = p \wedge p^c \ne 0$, because $p^c \in F$. Hence, first implication is correct.
Let F be a proper filter with the property that for all $p \in B$ for which $p \wedge q \ne 0$, for all $q \in F$ the $p$ is also in $F$. Let $p \in B$ be an arbitrary element. The $p$ either has the stated property or doesn't. If it does, the $p \in F$, hence $F$ is maximal. On the other hand, if there is a $q \in F$ such that $p \wedge q = 0$, does that imply that $q = p^c$? In that case $F$ would still be maximal, but I can't figure bit out.
Am I doing it right? Any help is appreciated.
For the second bullet point, what you really want to prove is that, for each $p \in B$, either $p \in F$ or $p' \in F$.
Now, if $p \notin F$, then, by the property that defines $F$, we have that $p \wedge q = 0$, for some $q \in F$, whence $q \leq p'$, yielding $p' \in F$.
Update.
About the question in your comment, from $p \wedge q = 0$ it follows that $q \leq p'$, using the well known Boolean laws: $$x \wedge y' = 0 \;\Longleftrightarrow\; x \leq y$$ and $x''=x$.
I'll prove directly the required implication.
If $p \wedge q = 0$, then \begin{align} q \wedge p' &= (q \wedge p') \vee 0\\ &= (q \wedge p') \vee (q \wedge p)\\ &= q \wedge (p' \vee p)\\ &= q \wedge 1 = q, \end{align} whence $q \leq p'$.