Thanks. But isn't there an embedded point at $t=0$? I will check this later.

Regarding 1: the ample cone is open, and its closure is the nef cone. Since the class in question is on the boundary, it cannot be ample.

I guess the right assumption that would make things work is that $f$ is proper and separable.

Very good example! Maybe one should add: we need boundedness of degree so that this "space of morphisms" is of finite type.

I don't get the question. Why not take $X = B\times C$ for a given genus-$g$ curve $C$?

In my example we have $f^{-1}(f(x)) = g^{-1}(g(x)) = x$ since $f$ and $g$ are both bijective...

Yes, as soon as the genus of $X$ is at least two.

What is $end$ ?

My guess is that this is impossible if the polarization of $J(X)$ is also defined over $K$, due to Torelli. So I would look for an abelian surface over $K$ with a principal polarization defined over $L$ that does not descend to $K$ (I think that any p.p. abelian surface is a Jacobian of a genus 2 curve).

Can you give a counterexample with $X$ and $Y$ algebraic?

I don't think that an abelian surface can be a complete intersection.

I think the right way to think about this is to take the trivial $l$-adic sheaf $\mathbb{Q}_l$ on $A$ and compute $G:=R^1\pi_* \mathbb{Q}_l$ -- a constructible $l$-adic sheaf on $C$. Then at every point $x$ of $C$ the action of the Frobenius on $G_x$ has characteristic polynomial whose roots are the Weil numbers corresponding to the fiber at $x$.