It does not say that A and B is mutually inclusive, so p(AUB) is between 1/2 and 1, right?

The easiest way to answer is to do a diagram of the situation you are facing

As you can see $P(A cap B)$ can be between zero and 0.5 (I indicated $A cap B$ as $AB$)

thus $0.5leq P(A cup B)leq1$

As said in comments, you are correct. Since $P(A)+P(B)-P(A∩B)=P(Acup B)$, we can calculate that $0le P(A∩ B)le frac{1}{2}$, and so $P(Acup B)$ is between $frac{1}{2}$ and $1.$