For instance,
"If $a \in \mathbb{Z},$ then $a^{5}>0$.
Now obviously the generalized conclusion is wrong because negative numbers belong to $\mathbb{Z}$ and a negative number to an odd integer power returns a negative number. However, $A \rightarrow B$ has some odd features about it.
$\begin{array}{c:c|c}A&B&A \to B\\\hline T&T& T\\ T&F& F\\ F&T& T\\ F&F& T\end{array}$
The conclusion as to whether a statement is true or false can't be interpreted from basic common sense, it has to be interpreted from a truth table of statements as that is the foundation for mathematical reasoning.
The hypothesis $a \in \mathbb{Z}$ may or may not be true depending on what you pick $a$ to be, so "if" $a$ belongs to natural numbers is already mystery, there are no other defined conditions besides what follows.
Then, it's obviously not always true that this implication $a^{5}>0$ is satisfied for $a$, but it is sometimes true. So, if A is true and B is false, then the statement is false, but the problem is it seems B could be either true or false depending on the specifics of the context.
The statement $$ 3\in \mathbb Z\to 3^5 >0$$ is true since the premise and conclusion are both true. The statement $$ -3\in \mathbb Z\to (-3)^5 > 0$$ is false since the premise is true and the conclusion is false. The statement$$ -\pi \in \mathbb Z \to (-\pi)^5 >0$$ is true since the premise is false and the conclusion is true.
Your problem is you aren't looking at a statement, you are looking at an open formula with a variable. One way to interpret what you mean by asking if the formula is true, is that you are actually asking about the universal closure $$ \forall a (a\in \mathbb Z \to a^5 > 0).$$ In this case, the sentence is false, because it's not true for all $a$ in the domain. (We haven't said what the domain is, but let's assume it's $\mathbb R$ for definiteness since both the predicates make sense and are not trivial there.) On the other hand, if you don't mean that, then we don't have enough information to answer the question since truth values of open formulas only make sense relative to a variable assignment.