An answer to (2): The algebra $\mathcal B$ is not only uncountable, it has the property that below any positive element there are uncountably many elements.
Let $\mathcal A'$ be the algebra of all sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where all $a_i,b_i$ are in $([0,\frac12]\cap \mathbb Q) \cup [\frac12,1]$. This is a subalgebra of $\mathcal A$. So $\mathcal A'/N$ is a subalgebra of $\mathcal A/N$, it is uncountable and atomless.
Now consider the element $[0,\frac12]/N$ in $\mathcal A'/N$. It is positive, but there are only countably many elements below it.
So $\mathcal A'/N$ is not isomorphic to $\mathcal B$.