Let me try to answer the modified question. (I assume we are working over an algebraically closed field of characteristic zero.)
The answer to your question depends crucially on what is meant by ruled. (I think this was a source of confusion in the comments.) There are two different defintions I know of, neither one standard:
- $S$ is geometrically ruled (what Hartshorne calls ruled) if $p: S \rightarrow C$ has every fibre isomorphic to $\mathbf{P}^1$;
- $S$ is birationally ruled (what some other people called ruled) if $S$ is birational to a product $C \times \mathbf{P}^1$.
Now Noether--Enriques tells you that a rational fibration is birationally ruled (at least once you know there are some smooth fibres, which you get from Bertini's second theorem) but you might worry that there are some singular fibres that stop it from being geometrically ruled. However, this can't actually happen, for the following reason. By your definition of rational fibration, every fibre of $p$ is an irreducible reduced rational curve, but any such curve which is not isomorphic to $\mathbf{P}^1$ must be singular, hence have arithmetic genus greater than 0; however, this cannot happen for a flat family such as $p: S \rightarrow C$. (See Hartshorne for justifications of all these assertions.)
So a rational fibration is the same thing as a geometrically ruled surface. Just to be sure that these are really different from the birationally ruled surfaces, let's give an example: take $p: S \rightarrow C$ a geometrically ruled surface, and let $\pi: S' \rightarrow S$ be the blowup of any point on $S$. Then $p \circ \pi: S' \rightarrow C$ is a birationally ruled surface, but one of its fibres is a reducible curve, so it is not geometrically ruled.
Actually, this last example also highlights a problem with your idea in the final paragraph: blowing up points unfortunately can't turn a fibration with singular fibres into one with smooth fibres, because it introduces extra components into some of the fibres.
I hope this helps.
Edit: Finally I think I can answer the intended question. Let $S$ be a birationally ruled surface, not isomorphic to $\mathbf{P}^2$. Contract $(-1)$-curves on $S$ until we reach a minimal surface $S_m$. So we have a morphism $p: S \rightarrow S_m$. Now according to Enriques' classification, there are three possibilities for $S_m$:
It is a geometrically ruled surface $\pi: S_m \rightarrow C$ over a curve of genus $>0$;
It is a Hirzebruch surace $\pi: S_m \rightarrow \mathbf{P}^1$;
It is $\mathbf{P}^2$.
The first two types are geometrically ruled, so $\pi \circ p$ gives a morphism to a curve with generic fibre $\mathbf{P}^1$. (We get $S$ from $S_m$ by blowing up a finite number of points, so that doesn't affect the generic fibre.)
The only problem is if we arrive at $\mathbf{P}^2$. But in that case we just stop contracting curves at the penultimate step, to get a morphism $p: S \rightarrow \Sigma_1$ where $\Sigma_1$ is the blowup of $\mathbf{P}^2$ in one point. Now $\Sigma_1$ is also a geometrically ruled surface, so we can argue as in the other cases.
And if that doesn't answer the question, I quit! :)
First,
$$
H^i(X,End(T_X)) = Ext^i(T_X,T_X).
$$
Second, there is a natural exact sequence
$$
0 \to T_{X/C} \to T_X \to p^*T_C \to 0,
$$
where $p:X \to C$ is the projection. Finally, we have relative Euler sequence
$$
0 \to O_X \to p^*E \otimes O_X(1) \to T_{X/C} \to 0,
$$
hence
$$
T_{X/C} \cong p^*\det(E) \otimes O_X(2).
$$
Combining all this and using long exact sequences of $Ext$'s one can compute the required cohomology groups.
Best Answer
The product topology (of Zariski topolgies) on $\mathbb{A}^1\times \mathbb{A}^1$ is not the same as the Zariski topology on $\mathbb{A}^2$, but topological product of Zariski topology is not what is usually meant by $\mathbb{A}^1 \times \mathbb{A}^1$, $\mathbb{P}^1 \times \mathbb{P}^1$ or $U_i \times \mathbb{P}^1$ in the definition of a $\mathbb{P}^1$ bundle. In Arapura, look at Proposition 2.4.6 for the definition of product of (pre)varieties.
The analytic topology on $\mathbb{A}^2$ is on the other hand the same as the product of the analytic topologies on $\mathbb{A}^1$ and $\mathbb{A}^2$ (and more generally for $X \times Y$).