EDIT: In my original post, I showed that $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof:
Theorem: $\mathfrak{x}_{lac} \leq \mathrm{non}(\mathcal N)$.
Note that this implies $\mathfrak{x}_{lac} < \mathrm{cov}(\mathcal N)$ in the random model, and it also implies the consistency of $\mathfrak{x}_{lac} < \mathfrak{d}$.
In addition to this theorem, let me also point out that it is consistent to have $\mathfrak{a} < \mathrm{cov}(\mathcal M)$. (See Corollary 2.6 in this paper of Brendle.) Therefore the lower bound $\mathrm{cov}(\mathcal M) \leq \mathfrak{x}_{lac}$ mentioned in the post already implies $\mathfrak{a}$ is not an upper bound for $\mathfrak{x}_{lac}$.
Proof of the theorem: Suppose we form an infinite set $B$ by choosing from each interval of the form $[2^k,2^{k+1})$ exactly one integer $b_k$ at random, and then taking $B = \{b_k :\, k \in \omega \}$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $[2^k,2^{k+1})$ has probability $1/2^k$ of being selected.) I claim that if $A$ is lacunary, then it is almost surely true that $A \cap B$ is finite.
To see this, fix some $c > 1$ and some $n_0 \in \omega$ such that if $a$ and $b$ are consecutive members of $A$ above $n_0$, then $b/a > c$. If $k$ is large enough that $n_0 < 2^k$, then this implies there are at most $log_c(2)$ members of $[2^k,2^{k+1})$ in $A$. This implies that the probability of choosing $b_k \in A$ is $\log_c(2)/2^k$ when $k$ is large enough. It follows that the probability of there being $>K$ members of $A$ in $B$ (when $K$ is large) is $\sum_{k > K} \log_c(2)/2^k = \log_c(2)/2^K$. Since this goes to $0$ as $K$ goes to infinity, the probability of $A \cap B$ being infinite is $0$.
The idea of choosing $B$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $b_k$'s. What the previous paragraph shows is that in this probability space, the set of all $B$'s with $A \cap B$ infinite form a null set, for any given lacunary set $A$. Hence any non-null subset of this probability space will contain a $B$ with $A \cap B$ finite. Since this holds for any $A$, we see that any non-null subset $X$ of this probability space contains, for any lacunary set $A$, some infinite $B$ with $A \cap B$ finite. The smallest possible size of such a set $X$ is $\mathrm{non}(\mathcal N)$.
QED
One more observation: The strict inequality $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $+ \, \neg \mathsf{CH}$, and then do a legnth-$\omega_1$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $\mathfrak{x}_{lac} = \aleph_1$ in the extension. But the iteration is $\sigma$-centered, and forcing with a $\sigma$-centered poset over a model of $\mathsf{MA}$ does not change the value of $\mathrm{non}(\mathcal N)$. Thus we get $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ in the extension.
Original post:
It is not provable that $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}_{lac}$, because $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model.
To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $[\omega]^\omega \cap V$ of infinite subsets of $\omega$ in the ground model has the property described in the definition of $\mathfrak{x}_{lac}$. That is: for every lacunary set $A \subseteq \omega$ in the extension, there is an infinite $B \subseteq \omega$ in the ground model such that $A \cap B$ is finite.
We work in the ground model. Suppose $\dot A$ is a name for an infinite lacunary set in the extension. There is some fixed $c > 1$ and $n_0 \in \omega$, and some forcing condition $p$, such that $p \Vdash$ if $a$ and $b$ are consecutive elements of $\dot A$ above $n_0$, then $b/a > c$.
I claim that for every $\varepsilon > 0$, there are arbitrarily large $n \in \omega$ such that $m(p \wedge [n \in \dot A]) < \varepsilon$.
(Note: Here I'm using the standard notation for forcing with measure algebras. If $\varphi$ is any well-formed formula in the forcing language, then we write $[\varphi]$ to mean the supremum of all the conditions forcing $\varphi$, and $m([\varphi])$ denotes the measure of this supremum. Roughly, you may think of $m([\varphi])$ as the probability that $\varphi$ ends up being true in the forcing extension.)
To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $\varepsilon > 0$ and some $N \in \omega$ such that $m([n \in \dot A]) \geq \varepsilon$ for all $n \geq N$. But this is just another way of saying that the "expected size" of $A \cap \{n\}$ is $\geq\varepsilon$ for all $n \geq N$. By the linearity of expectation, this means the expected size of $A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$ is $\geq (\ell-k)\varepsilon$ whenever $\ell > k > N$. But by our choice of $p$, if $k \geq N$ and $\ell < ck$, then $p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$. Since $c > 1$, this yields a contradiction for sufficiently large $k$, namely $k > N,2/c\varepsilon$.
Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $\dot A$ in the extension. Using the claim, we may find for each $k \in \omega$ some $n_k > n_{k-1}$ such that $m([n_k \in \dot A]) < m(p)/2^{k+2}$. Now let $p' = p - \bigvee_{k \in \omega}[n_k \in \dot A]$. Then $m(p') \geq m(p) - \sum_{k \in \omega}m([n_k \in \dot A]) > m(p)/2 > 0$, so $p'$ is a condition in our measure algebra, and $p'$ forces $\dot A$ to be disjoint from $\{n_k :\, k \in \omega \}$ (because it extends the complement of each $[n_k \in \dot A]$).
This shows that it is impossible to have a name $\dot A$ for a lacunary set such that $\dot A$ is forced to have infinite intersection with every infinite subset of $\omega$ from the ground model. Therefore there is no such set.
Yes, this is consistent.
Suppose we force to add $\kappa$ mutually generic Cohen reals to a model of $\mathsf{CH}$, where $\kappa$ is some cardinal with uncountable cofinality. In the extension, there are MAD families of cardinality $\aleph_1$ and cardinality $\kappa = \mathfrak{c}$, but there are no MAD families of any intermediate cardinality.
The proof of this is a basic instance of what's known as an "isomorphism of names" argument. I think this result appears as an exercise in Kunen's 1980 book, although I think it's originally due to Arnie Miller. The argument is greatly extended in
S. Shelah and O. Spinas, "MAD spectra," Journal of Symbolic Logic 80 (2015), pp. 243-262 (link).
They prove that the set of all cardinals $\kappa$ such that there is a MAD family of size $\kappa$ can be almost completely arbitrary. For example, given any $A \subseteq \omega \setminus \{0\}$, there is a forcing extension in which $A = \{ n < \omega :\, \text{there is a MAD family of size } \aleph_n\}$.
Best Answer
The cardinal $\mathfrak{nb}$ is equal to the reaping number $\mathfrak{r}$.
An unsplit family is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega$ with the property that $E \cap D$ and $E \setminus D$ are infinite for every $E \in \mathcal R$. The reaping number $\mathfrak{r}$ is defined to be the minimum cardinality of an unsplit family.
Note that a bipartite family (from the question) is defined in almost exactly the same way: just replace "are infinite" with "are nonempty" in the definition of an unsplit family.
The point is that $\mathcal E$ is a bipartite family then it is also an unsplit family, and if $\mathcal R$ is an unsplit family then $$\mathcal E = \{ X \subseteq \omega :\, X \Delta E \text{ is finite for some }E \in \mathcal R\}$$ is a bipartite family. Because neither kind of family can be finite, this implies that the minimum cardinality of a bipartite family, $\mathfrak{nb}$, is equal to the minimum cardinality of an unsplit family, $\mathfrak{r}$.