Fewest traversals to visit all vertices of DAG
I think you can solve this problem by reducing it to a minimum-cost circulation problem.

Let $G$ be your graph.  Replace each vertex $v$ with two vertices $v_\text{in},v_\text{out}$; replace each edge $u \to v$ into $v$ with an edge $u \to v_\text{in}$ and each edge $v \to w$ out of $v$ with an edge $v_\text{out} \to w$, and add the edge $v_\text{in} \to v_\text{out}$.    Add a new source vertex $s_0$ and a new sink vertex $t_0$.   For each vertex $v$ in your graph, add an edge $s_0 \to v_\text{in}$ and an edge $v_\text{out} \to t_0$.  Finally, add an edge $t_0 \to s_0$.  All edges have cost 0, lower bound 0, and upper bound $\infty$, except that all $v_\text{in} \to v_\text{out}$ edges have lower bound 1 and the $t_0 \to s_0$ edge has cost 1.

Now, find the minimum cost circulation in this graph.  There are polynomial-time algorithms to do this.

I claim that this minimum-cost circulation corresponds to the minimal set of paths to cover all the verticies.  In particular, if the minimum cost circulation has cost $c$, then that means there will be $c$ paths that cover the vertices.  This circulation will have total flow $c$, so you can trace where each of the $c$ units of flow go starting from $s_0$ to $t_0$, and that will trace out a path; take those $c$ paths as your collection of paths.  By the way I defined the lower bounds, this ensures that each vertex will be covered by at least one of these $c$ paths.