Permuting collinear points on a curve Let $C \subset {\bf CP}^2$ be an irreducible algebraic smooth (projectively) planar curve over the complex numbers of degree $d$ (we allow finitely many points to be deleted from $C$ to make it smooth).  Then a generic complex line in ${\bf CP}^2$ intersects $C$ in $d$ distinct points $z_1,\ldots,z_d$.

Question 1: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, is it possible to continuously deform the $d$ points $z_1,\ldots,z_d$ to $z_{\sigma(1)},\ldots,z_{\sigma(d)}$ while keeping all the points collinear, distinct, and on $C$ at all times?  In other words, does there exist continuous maps $z_i: [0,1] \to C$ with $z_i(0) = z_i$, $z_i(1) = z_{\sigma(i)}$, and the $z_1(t),\ldots,z_d(t)$ distinct and collinear for all $t\in [0,1]$?

I believe the answer to this question is yes, because one can transpose any two of the $z_i,z_j$ while keeping the other $z_k$ unchanged by bringing them close together near a generic point of $C$ (and making the collinear line close to the line of tangency of this point to $C$, which generically will not be tangent to anywhere else in $C$), performing the transposition, and then returning back to the original position.  One can also phrase the question as follows: if we let $S \subset C \times Gr(2,1)$ be the set of incidences $(p,\ell)$ with $p \in C$ and $\ell \in Gr(2,1)$ a line through $p$, then (after passing to an open dense subset of $Gr(2,1)$), $S$ is a covering space over (most of) $Gr(2,1)$ whose fibre has $d$ points, and the question asserts that the fundamental group of the base acts completely transitively on the fibre (in that every permutation of the fibre shows up); equivalently, the question asserts that the $d^{th}$ exterior power of $S$ over $Gr(2,1)$ is irreducible.
However, I'm having more trouble with proving the following generalisation of Question 1.  Keep the same setup as before, but now suppose we also have another irreducible curve $C'$ covering $C$ with fibres of cardinality $d'$.  Thus, above a generic $z_i$ in $C$ we have $d'$ points $w_{i,1},\ldots,w_{i,d'}$ in $C'$.

Question 2: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_d,k_1,\ldots,k_d \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d$ points $w_{1,j_1},\ldots,w_{d,j_d}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d),k_d}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct?

Like Question 1, one can phrase this question in terms of the transitivity properties of the action of the fundamental group of a suitable open dense subset of $Gr(2,1)$ on some fibre, but this formulation becomes rather complicated to state and I won't give it here.
I believe I can resolve this problem in the affirmative if $C$ has a smooth closure, in which case I can use ad hoc arguments to move around each of the $j_i$ independently, but encountered a problem if $C$ has a singular point in its closure, and $C'$ ramifies above this point, as then the points $z_i$ appear to become entangled with each other near this point in a manner that I was not able to analyse by ad hoc topological arguments.
[My motivation for this problem was to understand arcs in a finite plane ${\bf F}_p^2$, that is to say sets that do not contain any collinear triples, and to classify when these arcs can be constructed as the ${\bf F}_p$-points of a low-degree algebraic curve $C$ (or as the projection of the ${\bf F}_p$-points of a low-degree algebraic curve $C'$).  Using a Lefschetz principle argument, one can transfer the problem (in the large $p$ limit) to a problem in the complex plane related to the questions above.]
 A: While both of my original questions have been answered by David, I wanted to record the fact that a slightly weaker version of Question 2 is in fact true (and good enough for my particular application), in which one allows one of the points to move freely:

Question 2': for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_{d-1},k_1,\ldots,k_{d-1} \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d-1$ points $w_{1,j_1},\ldots,w_{d-1,j_{d-1}}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d-1),k_{d-1}}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct?

We sketch the proof as follows. Let $V \subset C^d$ be the set of all $k$-tuples of distinct collinear points $(z_1,\ldots,z_d)$ in $C$; by the positive solution to Q1, this is irreducible.  The embedding $V \subset C^d$ induces a homomorphism $\pi_1(V) \to \pi_1(C)^d$; by abstract nonsense, Q2' follows from (and is basically equivalent to) the assertion that the resulting image of $\pi_1(V)$ in $\pi_1(C)^d$ projects surjectively onto $\pi_1(C)^{d-1}$.
Now, since any two points determine a line, $C^2$ lifts to $V$, which implies that the projection of $\pi_1(V)$ to $\pi(C)^2$ is surjective.  Also, the image of $\pi_1(V)$ in $\pi_1(C)^d$ is symmetric with respect to permutations of the $d$ coordinates. This is enough to imply the surjectivity onto $\pi_1(C)^{d-1}$ by elementary group theory (this is a trick I learned from a paper of Furstenberg and Weiss).  Indeed, for any $g \in \pi_1(C)$, the surjectivity onto the final $\pi_1(C)^2$ implies that the image of $\pi_1(V)$ in $\pi_1(C)^d$ contains an element of the form $(g_1,\ldots,g_{d-1},g,1)$ for some $g_1,\ldots,g_{d-1}$; by symmetry it also contains $(g_1,\ldots,g_{d-1},1,g)$,and by dividing it thus contains $(1,\ldots,1,g,g^{-1})$.  Thus the projection onto $\pi_1(C)^{d-1}$ contains $(1,\ldots,1,g)$; permuting and composing we obtain the required surjectivity.
A: The answer to question 1 is yes, and is known as Harris's Uniform Position Lemma. It was proved in Harris's 1980 paper Galois groups of enumerative problems.
You can find a nice exposition in Chapter 9 of Solving Polynomial Equations, by Bronstein, Dickenstein and Emiris.
You should be warned that the analogous statement is not true in characteristic $p$, which you say is your motivation. For example, consider the curve $y=x^p$ in characteristic $p$. Any line $y=mx+b$ meets this curve at $p$ points, whose $x$ coordinates are the roots of $x^p=mx+b$. Note that the roots of this polynomial form an arithmetic progression: If $r$ and $s$ are roots, so is $r+k(s-r)$ for every $k \in \mathbb{Z}/p$. That means that the monodromy, as you move the line, must preserve this structure, so it is a subgroup of $\mathbb{Z}/p^{\ast} \ltimes \mathbb{Z}/p$. 

UPDATE I think question 2 is false! Let $C$ be a smooth cubic curve in $\mathbb{P}^2$, and choose a flex on $C$ to be the origin, so $C$ has a group structure. Take $C'$ to be equal to $C$, but with the map $C' \to C$ being the doubling map in the group law. So $d'=4$.
If $L$ is a line in $\mathbb{P}^2$, and $L \cap C = \{ x,y,z \}$, then $x+y+z=0$ in the group law. If $2 x'=x$, $2y' =y$ and $2 z'=z$, then $x'+y'+z'$ is a $2$-torsion point of $C'$. If we move $(L, x',y',z')$ continuously, we can't change which $2$-torsion point $x'+y'+z'$ is. So there are (at least) four separate connected components in the space of $(x',y',z')$ triples.
