Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite?

Yes. For every $r\in\mathbb R$ choose a sequence of rational numbers $\{r_n\in\mathbb Q\mid n\in\mathbb N\}$ which converges monotonically to $r$, this sequence is of course a subset of $\mathbb Q$ - a countable set.

If $r\neq s$ are two real numbers then the sequence we chose for them must intersect at a finite subset, otherwise we had a subsequence of the two which would converge to two different limit points.

Since $\mathbb R$ is uncountable (and in fact has cardinality as $\mathcal P(\mathbb Q)$) we have indeed uncountably many subsets of $\mathbb Q$ with the wanted property.

Yes. We know $\mathbb{N}$ and $\mathbb{Q}$ are equipotent so we choose a bijection $f:\mathbb{N} \to \mathbb{Q}$. We also know that $\mathbb{R}$ is equipotent to the set of equivalence classes of Cauchy sequences in $\mathbb{Q}$. For every $r \in \mathbb{R}$ choose $(q_{r,n})_n$ a representative from the equivalence class corresponding to $r$. Note that if $r_1\neq r_2 \in \mathbb{R}$ then $q_{r_1,n} = q_{r_2,n}$ for at most finitely many $n$. Since $f$ is a bijection we have that the sequences $(m_{r,n})_n := (f^{-1}(q_{r,n}))_n \subseteq \mathbb{N}$ share the same property. Since $\mathbb{R}$ is uncountable this concludes the proof; just choose the subset $N_\alpha$ to be the range of $m_\alpha$.

Let P be the set of all prime numbers, which is countable. Consider an infinite subset A = {$p_0,p_1,p_2...$} of primes from P arranged in an increasing order. We form $B_A$ = {$p_0,p_0p_1,p_0p_1p_2,...$} corresponding to A.

Now, the prime factorization is unique for integers. Thus, if we choose another infinite subset $A_1$ of P different from A then the corresponding $B_{A_1}$ and $B_A$ has only finitely many elements in common. Since P is countably infinite, the number of different infinite subsets of P is continuum. Hence we are done.