Four rays spread out from point $O$ in a $3$-dimensional space in a way that the angle between every two rays is $a$. Find $\cos a$.
1. **Understanding the problem**: We need to find the cosine of the angle \(a\) between any two rays emanating from a point \(O\) in 3-dimensional space such that the angle between every pair of rays is the same. This configuration suggests a regular tetrahedron, where the four rays correspond to the four vertices of the tetrahedron.

2. **Tetrahedral geometry**: In a regular tetrahedron, the angle between any two vertices (rays from the center to the vertices) is known as the tetrahedral angle. We need to find the cosine of this angle.

3. **Using geometric properties**: Consider a regular tetrahedron with vertices \(A, B, C, D\). The angle between any two vectors from the center \(O\) to the vertices is the same. We can use the dot product to find this angle.

4. **Vectors and dot product**: Let the vectors from the center \(O\) to the vertices be \(\vec{OA}, \vec{OB}, \vec{OC}, \vec{OD}\). The dot product of any two of these vectors gives us the cosine of the angle between them.

5. **Coordinates of vertices**: For simplicity, place the tetrahedron in a coordinate system such that the vertices are equidistant from the origin. One possible set of coordinates for the vertices of a regular tetrahedron centered at the origin is:
   \[
   A = \left(1, 1, 1\right), \quad B = \left(1, -1, -1\right), \quad C = \left(-1, 1, -1\right), \quad D = \left(-1, -1, 1\right)
   \]

6. **Calculating the dot product**: The dot product of any two vectors from the origin to these vertices is:
   \[
   \vec{OA} \cdot \vec{OB} = (1, 1, 1) \cdot (1, -1, -1) = 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot (-1) = 1 - 1 - 1 = -1
   \]

7. **Magnitude of vectors**: The magnitude of each vector is:
   \[
   |\vec{OA}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
   \]

8. **Cosine of the angle**: Using the dot product formula \(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta\), we have:
   \[
   \cos a = \frac{\vec{OA} \cdot \vec{OB}}{|\vec{OA}| |\vec{OB}|} = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3}
   \]

The final answer is \(\boxed{-\frac{1}{3}}\).