No. Take a triangle $v_0v_1v_2$ on the plane, $H'$ is its side $v_0v_1$, $H$ is almost another side $v_0v_2$. Then $\sum_{i=0}^2 {\rm dist}\, (v_i,H')$ is just a length of altitude from $v_2$, $\sum_{i=0}^2 {\rm dist}\, (v_i,H)$ is almost the length of altitude from $v_1$, which may be less than that from $v_2$.

No. Take a triangle $v_0v_1v_2$ on the plane, $H'$ is its side $v_0v_1$, $H$ is almost another side $v_0v_2$. Then $\sum_{i=0}^2 {\rm dist}\, (v_i,H')$ is just a length of altitude from $v_2$, $\sum_{i=0}^2 {\rm dist}\, (v_i,H)$ is almost the length of altitude from $v_1$, which may be less than that from $v_2$.

As for your new question, the answer is still negative for triangles. Assume that a median $v_0 p$ is perpendicular to $H'=v_0v_1$. For any line $H$ passing through $v_0$ and not cutting the triangle we have $\sum_{i=0}^2 {\rm dist}\,(v_i,H)=2\,{\rm dist}\, (p,H)\leqslant 2|v_0p|$ and maximum is achieved for $H=H'$.

No. Take a triangle $v_0v_1v_2$ on the plane, $H'$ is its side $v_0v_1$, $H$ is almost another side $v_0v_2$. Then both sides are (almost) lengths of altitudes, which may differ.

No. Take a triangle $v_0v_1v_2$ on the plane, $H'$ is its side $v_0v_1$, $H$ is almost another side $v_0v_2$. Then $\sum_{i=0}^2 {\rm dist}\, (v_i,H')$ is just a length of altitude from $v_2$, $\sum_{i=0}^2 {\rm dist}\, (v_i,H)$ is almost the length of altitude from $v_1$, which may be less than that from $v_2$.