If A is mapping reducible to B and A is recognizable, then is B recognizable too?
Answer for Corrected Question

Short answer, no.

You can think of reductions as "making problems harder" (somewhat inline with using $\leq$ as the base symbol for reductions of).

In this case we only have two levels of "hardness", a problem/language is either recognizable ("easy") or unrecognizable ("hard") (but this obviously extends to other degrees of computability and complexity).

So if $A \leq B$, and $B$ is "easy", $A$ must be "easy" too (because it's "less than" $B$). Contrapositively, if $A \leq B$ and $A$ is "hard, then $B$ must also be "hard" (because it's "greater than" $A$).

Using this intuition, it may be a bit easier to see why the latter two are wrong:


If $A \leq B$ and $A$ is "easy", this doesn't tell us much about $B$ (it could be "easy" or "hard").
Similarly, if $B$ is "hard", it doesn't tell us much about $A$.


Note that as with the first pair, the second are really the same statement, just logically inverted.

Old Answer, Just Here for Posterity

I think you may have misread something. The second (provided) theorem is not correct, it should be if $A \leq_{m} B$ and $A$ is unrecognizable, then $B$ is unrecognizable.

This is just the contrapositive of the first statement.

Following from this, the final two are mistaken.