The answer to Question 1 is NO:
Split $\omega$ into two disjoint infinite sets $X_0$ and $X_1$, fix $\mathcal A_0$ a countable AD family of subsets of $X_0$ such that
- $\mathcal A_0$ does not contain a $\Delta$-system of size $3$, yet
- for every finite $F\subseteq X_0$ there are $A,B\in \mathcal A_0$ such that $F=A\cap B$.
Added after comments below (To construct $\mathcal A_0$ start with a partition $\{B_n^{i}: n\in\omega, i\in 2\}$ of $\omega$ into infinite pieces and enumerate $[\omega]^{<\omega}$ as $\{F_n: n\in\omega\}$ . We shall recursively construct the family $\mathcal A_0=\{A_n^{i}:n\in\omega\}$ so that
(1) $A_n^{i}\subseteq^* B_n^{i}$,
(2) $F_n=A_n^0\cap A_n^1$ and
(3) $|A_n^{i}\cap A_m^{j}|\not = |A_n^{i'}\cap A_{m´}^{j'}|$ for every $m,m'<n$ and $i,i',j,j'\in 2$.
To accomplish this note that (1) gives you "enough room" to pick arbitrarily large disjoint pieces from the previous $A_m^{j}$'s which do not affect (2) to satisfy (3). Now, by (2) we have that $\mathcal A_0$ has the property that for every finite set $F\subseteq \omega$ there are $A,B\in \mathcal A_0$ such that $F=A\cap B$. The fact that we do not have "accidental" $\Delta$-systems of size $3$ follows directly from (3) as one (or two) of the lower indices of the potential 3-element $\Delta$-system come before the other and (3) guaranties that the corresponding intersections have different sizes.
Having done this just copy $\mathcal A_0$ on $X_0$).
Extend $\mathcal A_0$ to a maximal AD family $\mathcal A$ of subsets of $X_0$ not containing a $\Delta$-system of size $3$.
Claim. $\mathcal A$ is a maximal AD family of subsets of $\omega$ not containing a $\Delta$-system of size $3$ (yet not MAD).
$\mathcal A$ as a family of subsets of $\omega$ is not MAD as all elements of $\mathcal A$ are disjoint from $X_1$.
To show that it is maximal with respect to the property of not containing a $\Delta$-system of size $3$, assume that $B\in [\omega]^\omega$ AD with every element of $\mathcal A$. The proof has two simple cases:
Case 1: $C=B\cap X_0$ is infinite.
Then $C\cap A= B\cap A$ for every $A\in \mathcal A$ as $\mathcal A\subseteq \mathcal P(X_0)$. By maximality of $\mathcal A$ there are $A_0, A_1 \in\mathcal A$ such that $C\cap A_0=C\cap A_1= A_0\cap A_1$. Then, however, $B\cap A_0=B\cap A_1= A_0\cap A_1$.
Case 2: $C=B\cap X_0$ is finite.
Then by the property of $\mathcal A_0$ there are $A_0, A_1 \in\mathcal A$ such that $C= A_0\cap A_1$. Then $C=B\cap A_0=B\cap A_1= A_0\cap A_1$.
In both cases $\{B,A_0,A_1\}$ forms a $\Delta$-system.