$a)$ Let $l$ be the line through point $A$, such that $l\perp BC$. We will show that this is the locus. Indeed, as the height from $A$ to $BC$ is parallel to $l$, but also goes through point $A$, so the ortocenter of $\triangle ABC$ lies on $l$. Now, for every point $K$ on $l$, let the perpendicular from point $K$ to $p_{2}$ be $C_{1}$. Now $K$ is the orthocenter of $\triangle AC_{1}B_{1}$, so for every point on $l$ there exists a triangle, for which this point is it's orthocenter, so indeed, $l$ is the locus of those points.

$b)$ Similarly for the medians: If $G$ is the medicenter of $\triangle ABC$, then If $CG\cap AB = M_{C}$, then $CG:GM_{C}=2:1$, so $G$ lies on the line $d$, parallel to $AB$, such that $2dist(AB,d)=dist(d,c)$(but also $d$ is between the lines $AB$ and $c$), where $c$ is the line, which contains point $C$, parallel to $AB$. Now every point on $d$ can be a medicenter. Proof: Choose point $X\in d$ and now let $d_{1}$ be the line, such that $dist(AB,d_{1})=dist(d_{1},c)$. Let $AX\cap d_{1}=Y$. draw the line $B_{2}C_{2}$, such that $Y\in B_{2}C_{2}$ and $B_{2}C_{2}\parallel BC$ and $B_{2}, C_{2}$ are the intersections of that line with $AB$ and $c$ respectively. It's easy to show that $X$ is the medicenter of $\triangle AB_{2}C_{2}$.

$c)$ It's very similar to $a)$ and $b)$ so I'm just going to show how to construct the locus (line). If $a$ is a parallel line from $A$ to $c$ and let $a\cap c=B_{3}$. Now the midpoint of $B_{3}C_{3}$ is $M_{3}$. If $l\cap c=C_{4}$ and $M_{4}$ is the midpoint of $AB_{4}$ then the locus is the line $M_{3}M_{4}$.