Can you make a different learner?
Answer to questions 1 and 2: No. This is also the answer to 3 if $M$ is required to always halt (isn't this implied by the definition of a learner, or do you mean the program produced by $M$ must always halt?). Here's a proof for 1 (2 and 3 work pretty much the same way)

Suppose there's an algorithm $A$ which takes the description of a learner $N$ and outputs a learner $M$ which behaves differently. Let $N$ be a learner which runs $A$ on its own description, then runs the output $A(N) = M$ on the input text to produce a program that enumerates some set. By definition of $A$, $N$ and $M$ behave differently, but by construction they behave identically, contradicting the existence of $A$.

This same argument can be pretty easily modified to show that 2 is impossible. It can also be used to show that 3 is impossible if $M$ must always halt.

Not a real answer for question 4, but you might find this helpful: A prudent learner cannot converge on both every text for a finite set and on every text for the set of all natural numbers. Let $N$ be a learner that converges on every text for a finite set. Now, consider the following text for the set of all natural numbers. It contains all numbers in order, separated by differing numbers of 1s. After $i$, it contains enough 1s that $N$ will return a program which enumerates the set $1,\ldots,i$. $N$ will do this because it converges on every finite set. Because it will return a program enumerating a finite set after every prefix of this text, it does not converge on this text.

So if you let $M_1$ be the learner which converges on all finite sets and $M_2$ be the learner which converges on all texts for the set of all natural numbers, at least one of $M_1,M_2$ satisfies the criteria in question 4. You can extend this to a countably infinite set of learners in which at least all but one learner satisfies the criteria in question 4 by producing a machine $M_i$ for each natural number $i$ which converges on all finite sets which don't contain powers of the $i$th prime and also converges on the set of all powers of the $i$th prime. By similar logic as above, $N$ can't converge on all sets that any two of these converge on.

You can use this set of learners to find an $M$ satisfying 4 with arbitrarily high probability.