If $\mathcal B$ and $\mathcal B'$ are two bases for the same topology, then are $\mathcal B$ and $\mathcal B'$ comparable? 
Question 1)
  I know that if $\mathcal B \subset \mathcal B'$ then $\mathcal{T}_\mathcal{B} \subset \mathcal{T}_{\mathcal{B}'}$ but is the converse true? I.e. that if $\mathcal{T}_{\mathcal{B}} \subset \mathcal{T}_{\mathcal{B}'}$ then $\mathcal{B} \subset \mathcal{B}'$.
Question 2)
  If $\mathcal{T}$ has two differents bases $\mathcal{B}$ and $\mathcal{ B'}$, then does it hold that $\mathcal{B} \subset \mathcal{B}'$ and $\mathcal{B}' \subset \mathcal{B}$?

I  use the notation $\mathcal{B} \subset \mathcal{B}'$ to mean that if $B \in \mathcal{B}$ then there is $B' \in \mathcal{B}'$ such that $B'\subset B$.
 A: You need a different notion of "smaller", that also takes points in consideration:
For families of subsets of $X$ we define: 
$$\mathcal{B} \le \mathcal{B}' \text{ iff } \forall B \in \mathcal{B}: \forall x \in B: \exists B' \in \mathcal{B}': x \in B' \subseteq B$$
It's then easy to see that $\mathcal{B} \le \mathcal{B}'$ implies $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}_{\mathcal{B}'}$ because each member of $\mathcal{B}$ is open under $\mathcal{T}_{\mathcal{B}'}$.
From this it's easy to see that $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}_{\mathcal{B}'}$ also implies $\mathcal{B} \le \mathcal{B}'$. This implies a positive answer to the second question for this notion $\le$.
The OP's condition (I'll denote it $\le_P$ for Peter, say) 
$$\mathcal{B} \le_P \mathcal{B}' \text{ iff } \forall B \in \mathcal{B}: \exists B' \in \mathcal{B}': B' \subseteq B$$ does not necessarily imply that $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}_{\mathcal{B}'}$. This can be seen from $\mathcal{B} = \{[a,b) : a < b \in \mathbb{R}$ and $\mathcal{B}' = \{(a,b): a < b \in \mathbb{R}\}$, which obeys $\mathcal{B} \le_P \mathcal{B}'$ but the Sorgenfrey topology generated by $\mathcal{B}$ is strictly stronger (!) than the usual topology, which is generated by $\mathcal{B}'$. 
