It is given a regular hexagon in the plane. Let P be a point of the plane. Define s(P) as the sum of the distances from P to each side of the hexagon, and v(P) as the sum of the distances from P to each vertex.
a) Find the locus of points P that minimize s(P)
b) Find the locus of points P that minimize v(P)
### Part (a): Finding the locus of points \( P \) that minimize \( s(P) \)

1. **Define the problem in terms of geometry and area**:
   Let \( A_1, A_2, \ldots, A_6 \) be the vertices of the regular hexagon, and let \( a \) be the side length of the hexagon. The function \( s(P) \) is defined as the sum of the distances from a point \( P \) to each side of the hexagon.

2. **Express the sum of distances in terms of area**:
   The sum of the distances from a point \( P \) to the sides of the hexagon can be related to the area of the hexagon. Specifically, if \( d_i \) is the distance from \( P \) to the \( i \)-th side, then:
   \[
   s(P) = d_1 + d_2 + \cdots + d_6
   \]
   The area of the hexagon can be expressed as:
   \[
   \text{Area} = \frac{3\sqrt{3}}{2} a^2
   \]
   Since the sum of the distances from any point \( P \) inside the hexagon to its sides is constant and equal to the height of the hexagon, which is \( \frac{3\sqrt{3}}{2} a \), the minimum value of \( s(P) \) is achieved when \( P \) is inside the hexagon.

3. **Conclusion**:
   Therefore, the locus of points \( P \) that minimize \( s(P) \) is the interior of the regular hexagon.

\[
\boxed{\text{The locus of points } P \text{ that minimize } s(P) \text{ is the interior of the regular hexagon.}}
\]

### Part (b): Finding the locus of points \( P \) that minimize \( v(P) \)

1. **Define the problem in terms of the complex plane**:
   Place the hexagon in the complex plane such that its circumcircle is the unit circle centered at the origin. Let the vertices of the hexagon be \( A_1, A_2, \ldots, A_6 \).

2. **Use the triangle inequality**:
   The function \( v(P) \) is defined as the sum of the distances from a point \( P \) to each vertex of the hexagon. Let \( P \) be represented by a complex number \( z \), and the vertices \( A_i \) be represented by complex numbers \( z_i \) on the unit circle. Then:
   \[
   v(P) = |z - z_1| + |z - z_2| + \cdots + |z - z_6|
   \]
   By the triangle inequality, the sum of the distances from \( P \) to the vertices is minimized when \( P \) is at the center of the hexagon, which is the origin in the complex plane.

3. **Conclusion**:
   Therefore, the locus of points \( P \) that minimize \( v(P) \) is the center of the hexagon.

\[
\boxed{\text{The locus of points } P \text{ that minimize } v(P) \text{ is the center of the hexagon.}}
\]