Solution:

(a) The points $P$ for which $l(P)$ is minimal are the points interior to the hexagon and those on its boundary. Let us call $A B C D E F$ the hexagon, and $a$ the length of its side, and let us consider the sum of the distances between $P$ and the two opposite sides $A B$ and $D E$. If $H$ and $K$ are the feet of the perpendiculars from $P$ to the lines $A B$ and $D E$ respectively, then, by definition, $P H$ and $P K$ are the distances of $P$ from the lines $A B$ and $D E$. $P, H$ and $K$ are collinear, because $P H$ and $P K$ are both perpendicular to the two lines, and $H K$ is equal to the distance between the two lines, that is, to $\sqrt{3} a$.

If $P$ lies inside the segment $H K$ (endpoints included) then $H P+P K=H K=\sqrt{3} a$. If instead it is outside, then the longer of the two segments $P H$ and $P K$ contains within it the entire segment $H K$, so in particular $P H+P K>H K>\sqrt{3} a$. It is evident that the first case occurs when $P$ is interior to the strip formed between the two lines $A B$ and $D E$ (boundary included), while the second occurs when $P$ is exterior to that strip. In conclusion,
$$\operatorname{dist}(P, A B)+\operatorname{dist}(P, D E)=P H+P K \geqslant \sqrt{3} a$$
and equality holds if and only if $P$ is interior to the strip (boundary included). Repeating the reasoning for the other two pairs of opposite sides, we get
$$\begin{aligned} & \operatorname{dist}(P, A B)+\operatorname{dist}(P, D E) \geqslant \sqrt{3} a, \\ & \operatorname{dist}(P, B C)+\operatorname{dist}(P, E F) \geqslant \sqrt{3} a, \\ & \operatorname{dist}(P, C D)+\operatorname{dist}(P, F A) \geqslant \sqrt{3} a, \end{aligned}$$
and equalities hold respectively if $P$ is interior to the three strips determined by the pairs of parallel lines $\{A B, D E\},\{B C, E F\}$, $\{C D, F A\}$ (boundaries included).

Adding the three relations found, we obtain

$$\operatorname{dist}(P, A B)+\operatorname{dist}(P, B C)+\operatorname{dist}(P, C D)+\operatorname{dist}(P, D E)+\operatorname{dist}(P, E F)+\operatorname{dist}(P, F A) \geqslant 3 \sqrt{3} a .$$
Equality holds when $P$ is in the intersection of the three strips formed by the pairs of opposite sides of the hexagon, that is, when $P$ is inside the hexagon or on its boundary.

(b) The only point for which $v(P)$ is minimal is the center of the hexagon.

To prove this, let us consider the sums of the distances of $P$ from the pairs of opposite vertices $\{A, D\},\{B, E\}$, $\{C, F\}$ ). By the triangle inequality, $A D \leqslant A P+P D$, and equality holds if and only if $P$ is on the segment $A D$. Similarly, $B E \leqslant B P+P E$, with equality only if $P$ lies on the segment $B E$, and $C F \leqslant C P+P F$, with equality only if $P$ lies on the segment $C F$. Adding the three relations we obtain
$$\begin{aligned} v & =P A+P B+P C+P D+P E+P F \\ & =(A P+P D)+(B P+P E)+(C P+P F) \leqslant A D+B E+C F=6 a \end{aligned}$$
and equality holds if and only if the equality conditions of the three relations we added hold simultaneously, that is, if $P$ lies simultaneously on the segments $A D, B E$ and $C F$. It is evident that the only point satisfying this condition is the center of the hexagon.