Here is a complete proof: as remarked in the answer by Norondion, we can reduce to
the case when $C\_1 \rightarrow C\_2$ is generically separable, i.e. $k(C\_1)$ is separable
over $k(C\_2)$. Let $A \subset k(C\_1)$ be a finite type $k$-algebra consisting of the regular
functions on some non-empty affine open subset $U$ of $C\_2$ (it doesn't matter which one
you choose), so that $k(C\_2)$ is the fraction field of $A$.
By the primitive element theorem, we may write $k(C\_1) = k(C\_2)[\alpha]$, where
$\alpha$ satisfies some polynomial $f(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1} + \cdots
+ a_1 \alpha + a_0 = 0,$ for some $a_i$ in $K(C\_2)$.
Now the $a_i$ can be written as fractions involving elements of $A$, i.e. each
$a_i = b_i/c_i$ for some $b_i,c_i \in A$ (with $c_i$ non-zero). We may replace
$A$ by $A[c\_0^{-1},\ldots,c\_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at
the zeroes of the $c_i$), and thus assume that in fact the $a_i$ lie in $A$.
The ring $A[\alpha]$ is now integral over $A$, and of course has fraction field equal
to $k(C_2)[\alpha] = k(C_1)$. It need not be that $A[\alpha]$ is integrally closed,
though. We are going to shrink $U$ further so we can be sure of this.
By separability of $k(C_1)$ over $k(C_2)$, we know that the discriminant $\Delta$
of $f$ is non-zero, and so replacing $A$ by $A[\Delta^{-1}]$ (i.e. shrinking $U$
some more) we may assume that $\Delta$ is invertible in $A$ as well.
It's now not hard to prove that $A[\alpha]$ is integrally closed over $A$. Thus
$\text{Spec }A[\alpha]$ is the preimage of $U$ in $C_1$ (in a map of smooth curves,
taking preimages of an affine open precisely corresponds to taking the integral closure
of the corresponding ring).
In other words, restricted to $U \subset C_2$, the map has the form
$\text{Spec }A[\alpha] \rightarrow \text{Spec }A,$ or, what is the same,
$\text{Spec }A[x]/(f(x)) \rightarrow \text{Spec A}$.
Now if you fix a closed point $\mathfrak m \in \text{Spec }A,$ the fibre over this point
is equal to $\text{Spec }(A/\mathfrak m)[x]/(\overline{f}(x)) = k[x]/(\overline{f}(x)),$
where here $\overline{f}$ denotes the reduction of $f$ mod $\mathfrak m$.
(Here is where we use that $k$ is algebraically closed, to deduce that $A/\mathfrak m = k,$
and not some finite extension of $k$.)
Now we arranged for $\Delta$ to be in $A^{\times}$, and so $\bar{\Delta}$ (the reduction
of $\Delta$ mod $\mathfrak m$, or equivalently, the discriminant of $\bar{f}$)
is non-zero, and so $k[x]/(\bar{f}(x))$ is just a product of copies of $k$,
as many as equal to the degree of $f$, which equals the degree of $k(C_1)$ over
$k(C_2)$. Thus $\text{Spec }k[x]/(\bar{f}(x))$ is a union of that many points,
which is what we wanted to show.
Yes. In fact what Artin proves in SGA4 exp XI thm 4.4 is that étale cohomology and singular cohomology agree for smooth schemes over $\mathbb{C}$ with finite coefficients. The statement you want will follow from this by taking inverse limits to get to $\mathbb{Z}_\ell$ and then extending scalars to $\mathbb{Q}_\ell$. If you don't feel like looking at SGA, you can find treatments of this in the books by Freitag-Kiehl, Milne,…
Added (in response to comment). The isomorphism $H_{et}^*(X_{\bar K}, \mathbb{Q}_\ell)\cong H_{et}^*(X_{\mathbb{C}}, \mathbb{Q}_\ell)$ follows from the smooth base change theorem (cf. Milne, Etale cohomology, p 231 cor 4.3).
Best Answer
See this explanation http://therisingsea.org/notes/Section2.6-Divisors.pdf Proposition 20.