Does showing a problem and its complement are not Turing-decidable means that the language & its complement are not Turing-recognizable?
... a problem and its complement are not Turing-decidable means that the language &amp; its complement are not Turing-recognizable?  


This claim is false. In fact, at least one of them (or both are) is Turing-recognizable. Assume that both the set (problem) $A$ and its complement $\overline{A}$ are not decidable. Then if both $A$ and $\overline{A}$ were recognizable (recursively enumerable) then that would mean $A$ and hence $\overline{A}$ are decidable (Theorem 4.22), a contradiction.   


  ... however, the Theorem 5.30 states something different.


This theorem states that $EQ_{TM}$ is neither Turing-recognizable nor co-Turing-recognizable. In other words, both $EQ_{TM}$ and $\overline{EQ}_{TM}$ are not Turing-recognizable (recursively enumerable). This does not contradict to the above-stated claim.

Also, pay attention that the theorem 5.30 reduces $A_{TM}$ to $\overline{EQ}_{TM}$ (not to $EQ_{TM}$ as you claim in your post)  meaning that $EQ_{TM}$ cannot be Turing-recognizable (otherwise $\overline{A}_{TM}$ would be Turing-recognizable and hence decidable).      

You are right, if a language is not decidable, then either the language itself, its complement or both of them are not Turing-recognizable, and since the theorem 5.30 proves that $EQ_{TM}$ is not decidable, it follows that either $EQ_{TM}$ either, $\overline{EQ}_{TM}$ or both of them are not Turing-recognizable. But this result is weaker and does not contradict to the theorem 5.30. In fact the latter (weaker) result is not necessary since the theorem already establishes the stronger result.