Let's do some vector algebra

Consider the cross product:

$\vec{a} \times \vec{c} = \vec{d} = \lvert \vec{a} \rvert \lvert \vec{c} \rvert \sin{\theta} \ \hat{d} = 2 \sin{\theta} \ \hat{d}$

Where $\hat{d}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{c}$ . Now consider:

$\vec{a} \times (\vec{a} \times \vec{c} ) = -\vec{b}$

$\implies 2 \sin{\theta} \ (\vec{a} \times \hat{d}) = -\vec{b}$

Since $\vec{a}$ and $\hat{d}$ are mutually perpendicular unit vectors, their cross product must also be a unit vector which is parallel to the unit vector $\vec{b}$ . It may be in the same or opposite direction to $\vec{b}$ however. This leads to:

$\pm \ 2 \sin{\theta} \ \vec{b} = -\vec{b}$

$\implies \lvert \pm 2 \sin{\theta} \rvert = 1$

$\implies \theta = \arcsin\left(\frac{1}{2}\right) =30^{\circ}$

The magnitude of the angle between the vectors $\vec{a}$ and $\vec{c}$ is $\boxed{30^{\circ}}$

$\vec a\times ( \vec a\times \vec c) +\vec b=0\implies \vec a(\vec a. \vec c) -a^2\vec c+\vec b=0\implies \vec b=a^2\vec c-\vec a(\vec a. \vec c) \implies b^2=\left (a^2\vec c-\vec a(\vec a. \vec c) \right ) ^2$ .

Substituting magnitudes we get

$4+(\vec a. \vec c) ^2-2(\vec a. \vec c) ^2=1\implies (\vec a. \vec c) ^2=3$ .

Since we are asked for the acute angle, we get for the required angle $α$ :

$α=\cos^{-1} \frac{\vec a. \vec c}{ac}=\cos^{-1} \frac{\sqrt 3}{2}=\boxed {30\degree}$ .