I can answer problems 2 and 3, although I still don't know the answer to problems 0 and 1.
The main point is that

$\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model.

I'll sketch a proof of this below. (It's a bit long, but I've tried to make it readable.) The proof actually shows a little more: it gives you $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$.

This result also answers problem 3, because we know that $\mathfrak{z} = \aleph_1$ in the random model. (This was first proved by Alan Dow and David Fremlin here. It is also a corollary to Theorem 4.2 in this paper by me and Alan.) Therefore $\mathfrak{z} < \mathfrak{j}_{2:2},\mathfrak{j}_{2:1}$ is consistent. On the other hand, Koppelberg proved that $\mathfrak{z} \leq \mathrm{cov}(\mathcal{M})$. (Actually, she proved the dual statement in the category of Boolean algebras here. Stefan Geschke wrote a purely topological proof here.) Because you have proved that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} \leq \mathrm{non}(\mathcal{M})$, and because $\mathrm{non}(\mathcal{M}) < \mathrm{cov}(\mathcal{M})$ in the Cohen model, it follows that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} < \mathfrak{z}$ is consistent. Thus there is no inequality between $\mathfrak{z}$ and either of $\mathfrak{j}_{2:2}$ of $\mathfrak{j}_{2:1}$ that is provable in $\mathsf{ZFC}$.

(I know I gave a different argument for this in the comments. I don't like that argument as much because it relies on Alan's unpublished -- and mostly unwritten -- argument that $\mathfrak{z} = \aleph_1$ in the Laver model. I'm sure he's right. But I like that the argument here relies on the fact that $\mathfrak{z} = \aleph_1$ in the random model, and you can go read one or two proofs of this if you like.)

Now let's sketch the proof that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model. For the sake of clarity, I'm going to avoid forcing jargon and give a probabilistic argument that (I hope) will give you the right idea.

To show that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model, let's first recall how random real forcing works. Roughly, we imagine ourselves to live in a universe $V$ of sets, containing real numbers, subsets of $\mathbb N$, lots of $2$-to$1$ functions, and whatever else. But we know that our universe is about to get bigger -- this is the forcing -- by the introduction of a "truly random" real number $r$. The new, bigger universe is called $V[r]$.

The first observation I'd like to make is that all continuous measures on uncountable Polish spaces are essentially isomorphic. This means that it doesn't matter whether we view $r$ as a random element of $\mathbb R$, or of $[0,1]$, or of $2^\omega$ with the standard product measure, or whatever. For this problem, we want to view $r$ as an infinite sequence of random selections from larger and larger finite sets $I_n$, where $I_n$ has size $n!$. We select, at random, only a single element from each set. (This can be formalized by saying that we'd like $r$ to be a random element of the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$, equipped with the usual product measure. But let's keep it informal.) So our universe is about to get bigger by introducing a truly random sequence of selections from some sets $I_0, I_1, I_2, \dots$ with $|I_n| = n!$.

Within $V$, we can try to anticipate objects that will be constructible from $r$ in $V[r]$. For example, we can anticipate that once we get $r$, we can build a set $J \subseteq \mathbb N$ according to the following recipe: first identify $I_n$ with the interval $[1+1+2+\dots+(n-1)!,1+1+2+\dots+(n-1)!+n!) \subseteq \mathbb N$, and then let the $n^{\mathrm{th}}$ element of $J$ be whatever $r$ randomly selects from this interval.

Now I claim that this set $J$ described above has the following property: if $f$ is any $2$-to-$1$ function in the ground model $V$, then the restriction of $f$ to $J$ is almost-injective. To prove this, it suffices to argue that it's true with probability $1$, given that $r$ makes its selections randomly. This suffices because this is precisely what we mean when we say that $r$ is a "truly random" addition to $V$: if there is a randomness test defined in $V$ (such as one defined from any $f \in V$), then $r$ is random with respect to that test.

So let's argue probabilistically.
Fix a $2$-to-$1$ function $f \in V$.
If $f(a) = f(b)$, we may view this as a "guess" that $f$ is making about our set $J$: the guess is that $a$ and $b$ are both in $J$. In other words, $f$ gets to guess at pairs from $J$ infinitely many times, and it is our job to prove that, with probability $1$, only finitely many of these guesses are correct.

So what is the probability that $f$ correctly guesses a pair of elements from $J$?
If $f$ identifies a member of some $I_m$ with a member of some $I_n$, where $m \neq n$, then there is a probability of exactly $\frac{1}{m!n!}$ that $f$ will have correctly guessed a pair from $J$. When $f$ makes other kinds of guesses (not identifying some member of some $I_m$ with a member of some $I_n$, where $m \neq n$), then the probability is $0$ that $f$ will have correctly guessed a pair from $J$.

If $m < n$, then $f$ gets at most $|I_m| = m!$ chances to guess a pair from $J$ with one member in $I_m$ and the other in $I_n$. By the previous paragraph, the probability of one of these guesses being correct is $\leq\! m!\frac{1}{m!n!} = \frac{1}{n!}$. Summing over all $n > m$, it follows that the probability of $f$ correctly guessing any pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m$ is
$$\leq\! \sum_{n = m+1}^\infty \frac{1}{n!} < \frac{1}{(m+1)!} \sum_{k = 0}^\infty \frac{1}{(m+1)^k} = \frac{1}{(m+1)!}\frac{m+1}{m} < \frac{1}{m!m}.$$

Now fix $k > 0$. Summing over all $m > k$, we see that the probability of $f$ correctly guessing a pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m > k$ is
$$\leq\! \sum_{m = k+1}^\infty \frac{1}{m!m} < \sum_{m = k+1}^\infty \frac{1}{m!} < \frac{1}{k!k}.$$

Therefore the probability of $f$ correctly guessing a pair of elements from $J \setminus (I_0 \cup \dots \cup I_k)$ is at most $\frac{1}{k!k}$. For any fixed $\varepsilon > 0$, we can choose $K$ large enough that $\sum_{k = K}^\infty \frac{1}{k!k} < \varepsilon$. This means that for $K$ large enough, the probability of $f$ correctly guessing more than ${K+1} \choose 2$ pairs of elements of $J$ is less than $\varepsilon$. Therefore the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is less than $\varepsilon$. As $\varepsilon$ was arbitrary, the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is $0$.

This shows that our set $J$ in $V[r]$ "should" (probabilistically) have the property that $f \restriction J$ is almost injective for every $f \in V$. But as we said earlier, this means $J$ really does have this property.

Why does this mean $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model? The random model is $V[G]$, where $G$ is a "random" element of the measure algebra $2^{\aleph_2}$. If $\mathcal F$ is any set of $\aleph_1$ $2$-to-$1$ functions in $V[G]$, then a standard "nice names" argument shows that there is some weight-$\aleph_1$ subalgebra $X$ of $2^{\aleph_2}$ such that $\mathcal F$ is already in the intermediate model $V[X \cap G]$. Because $|X| = \aleph_1$, there will be random reals added in moving from the intermediate model $V[X \cap G]$ to the final model $V[G]$ -- random over $V[X \cap G]$, not just over $V$. We've just showed that the addition of these random reals adds some $J$ that "works" for every $f \in \mathcal F$.

Why does this mean $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$? There are a few ways to see this. The easiest is probably just to go through the above argument and convince yourself that what we've really proved is that every $2$-to-$1$ function $f$ is "solved" by a measure-$1$ set of $J$'s in the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$. Equivalently, the set of $J$'s that fail to work for a given $2$-to-$1$ function $f$ is a null set $N_f$.
Therefore, if $\mathcal F$ is any size $<\! \mathrm{cov}(\mathcal{N})$ family of $2$-to-$1$ functions, $\bigcup_{f \in \mathcal F}N_f$ does not cover our Polish space, and so there is some $J$ that works for every $f \in \mathcal F$.