Answer
The angle between the two vectors is either $36.8^{\circ}$ or $143.2^\circ$.
Work Step by Step
The magnitude of the vector product $\vec{A} \times \vec{B}$, represented as $\vert \vec{A} \times \vec{B} \vert$, is given by
$$ \vert \vec{A} \times \vec{B} \vert = AB \sin \phi
$$ where $A$ and $B$ are the magnitudes of $\vec{A}$ and $\vec{B}$, respectively, and $\phi$ is the angle between them. From this, we can find $\phi$ as
$$ \phi = \sin^{-1} \left( \dfrac{\vert \vec{A} \times \vec{B} \vert}{AB} \right)
$$ We can find $\vert \vec{A} \times \vec{B} \vert$ as follows:
$$\vert \vec{A} \times \vec{B} \vert = \sqrt{(-5.00)^2+(2.00)^2} = \sqrt{29.00}
$$ Since $A = 3.00$ and $B = 3.00$, then
$$ \phi = \sin^{-1} \left( \dfrac{\sqrt{29.00}}{\left(3.00\right)\left(3.00\right)} \right) = \sin^{-1} \left( 0.598351645\ldots \right)
$$ and this result has two values. One of them is $36.8^\circ$, rounded off to the nearest tenths. The other one is the supplement of this angle, $180^\circ - 36.8^\circ = 143.2^\circ$. We can check this by showing that
$$ \sin \left( 36.8^\circ \right) = \sin \left( 143.2^\circ \right)
$$ Thus the angle between the two vectors is either $36.8^{\circ}$ or $143.2^\circ$.
