Donu Arapura is right:

If you want $F$ isomorphic to $E$ you can do the following:
Consider the product $E \times P^1$. You find $F \simeq Alb(X) \simeq E$.

If you want $F$ and $E$ to be non isomorphic, one can do as follows.
Let $c$ be a 2-torsion point of $E$. Form the quotient $X$ of $E \times E$ by the involution defined by $(x, y) \mapsto (-x, y + c)$.

The first projection induced an elliptic fibration $X \to \mathbb P^1$. A generic fibre is $E$.
However, the albanese of $X$ is $E'$ = $E$/{0, $c$}.

The statement about the albanese follows by considering the elliptic fibration $X \to E'$ given by the second projection, and the statement of Beauville.

So let us continue this discussion with examples of Kodaira dimension one:

Take an elliptic curve $E$ and a curve $C$ with $g(C) \ge 2$.
Then $E \times C \to C$ is an example with $E$ and $F$ being isomorphic.

Now let $C$ be an hyperelliptic curve with involution $i \colon C \to C$.
Let $E$ be elliptic, and let $b \in E$ be a 2-torsion point.
We define $X$ to be the quotient of $E \times C$ by the involution given by $(x, y) \mapsto (x + b, i(y))$.
As before, the fibration $X \to \mathbb P^1$ gives an example for $E$ not isomorphic to $F$.