As mentioned in other answers, the angle between two vectors (in 2D or 3D) is usually measured in $[0°, 180°]$. However, if we choose to measure positive angles counterclockwise, and negative angles counterclockwise, we can extend the range to $[-180°, +180°]$,
I shall assume a right-handed coordinate system, and explain the details, in this answer.
In two dimensions, the analog for cross product can be used to determine if the second vector is to the left (positive angles) or to the right (negative angles) of the first vector. This essentially expands the angle to range $(-180°, +180°)$, where positive angles are measured counterclockwise, and negative angles clockwise.
If we have vectors $\vec{u} = ( x_u, y_u )$ and $\vec{v} = ( x_v, y_v )$, then
$$\begin{cases}
\cos\varphi = \frac{\vec{u} \, \cdot \, \vec{v}}{\left\lvert\vec{u}\right\rvert \, \left\lvert\vec{v}\right\rvert} \\
\sin\varphi = \frac{\vec{u} \, \times \, \vec{v}}{\left\lvert\vec{u}\right\rvert \, \left\lvert\vec{v}\right\rvert}
\end{cases} \; \Leftrightarrow \; \begin{cases}
\cos\varphi = \frac{ x_u x_v + y_u y_v }{\sqrt{(x_u^2 + y_u^2)(x_v^2 + y_v^2)}} \\
\sin\varphi = \frac{ x_u y_v - y_u x_v }{\sqrt{(x_u^2 + y_u^2)(x_v^2 + y_v^2)}}
\end{cases}$$
However, if you already know $\cos\varphi$, then it is sufficient to check the sign of $x_u y_v - y_u x_v$, as $\sin\varphi$ will have the same sign, and value $\;\pm\sin\varphi = \sqrt{1 - \cos^2\varphi}$. This also means that
$$\begin{cases}
\varphi = \pm\arccos\left( \frac{ x_u x_v + y_u y_v }{\sqrt{(x_u^2 + y_u^2)(x_v^2 + y_v^2)}} \right ) & \, \\
\varphi < 0, & x_u y_v - y_u x_v < 0 \\
\varphi > 0, & x_u y_v - y_u x_v > 0
\end{cases}$$
Note that the above leaves the $x_u y_v - y_u x_v = 0$ case undefined. In this case, $\varphi = \pm 180°$, and it is up to you to choose the sign.
In three dimensions, vector cross product yields a vector, and is therefore not as directly usable as the 2D analog is. (In particular, the length of the result vector is always positive, so if you tried to extend the above to 3D, you'd always get a positive $\sin\varphi$).
However, if you have two pairs of vectors on the same plane, you can compare their orientation or order very easily.
Let's say you have a prototype vector pair, where the second vector is counterclockwise from the first, and their vector cross product is $\vec{n}$, $\left\lvert\vec{n}\right\rvert \ne 0$. (As long as its length is not zero, its length is irrelevant, only the direction matters. That is also why all the vectors need to be on the same plane.) Then, with $\vec{u}$ and $\vec{v}$ on the same plane as the prototype vector pair,
$$\begin{cases}
\left(\vec{u} \times \vec{v}\right) \cdot \vec{n} < 0, & \vec{v} \; \text{is clockwise from} \; \vec{u} \\
\left(\vec{u} \times \vec{v}\right) \cdot \vec{n} > 0, & \vec{v} \; \text{is counterclockwise from} \; \vec{u}
\end{cases}$$
If $\vec{u} = -\vec{v}$ ($\left(\vec{u} \times \vec{v}\right) \cdot \vec{n} = 0$), the above leaves the "direction" undecided; it's up to you to decide whether it means $-180°$ or $+180°$.
In general in 3D, if you have a third vector, say $\vec{w}$, that is perpendicular to both $\vec{u}$ and $\vec{v}$, you can use it to define the "handedness". Assuming right-handed coordinate system, we have
$$\begin{cases}
\left(\vec{u} \times \vec{v}\right) \cdot \vec{w} < 0, & \vec{v} \; \text{is clockwise from} \; \vec{u} \\
\left(\vec{u} \times \vec{v}\right) \cdot \vec{w} > 0, & \vec{v} \; \text{is counterclockwise from} \; \vec{u}
\end{cases}$$
The lack of a generally applicable $\vec{w}$ is the reason why the 2D case cannot be simply extended to 3D. We could rotate $\vec{u}$ and $\vec{v}$ so that $\vec{u}$ would be on the positive $x$ axis, and $\vec{v}$ on the $z=0$ plane; but, we do not have enough information to decide whether the rotation should rotate $\vec{v}$ to positive or negative $y$ coordinates. Essentially, in 3D, the clockwise/counterclockwise orientation depends on which side of the plane formed by the two vectors and origin you look from!
Best Answer
Hint: Calculate
$\displaystyle \cos \theta = \frac{a . b}{|a||b|}.$
What do you get?
$\displaystyle a . b = \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + (0) = \frac{\sqrt{6}}{3} = \sqrt{\frac{2}{3}}$
$\displaystyle |a| .|b| = \left|~\sqrt{\left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2 }~\right|. \left|~\sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 +0^2}~\right| = |1|.|1| = 1$
This comes out to: $\displaystyle \cos^{-1} \sqrt{\frac{2}{3}}$