does a line bundle always have a degree

One generalization of degree is first Chern class: A Cartier divisor corresponds to a class in $H^1(X;\mathcal{O}_X^{\times})$, and you take its image under the boundary map of the long exact sequence corresponding to the exponential exact sequence $\mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^{\times}$ where the second map is taking exponential (if you want to work in the algebraic category, there is a fix for this, using the exact sequence $\mathbb{Z}/n\mathbb{Z} \to \mathcal{O}_X^{\times} \to \mathcal{O}_X^{\times}$, where the second map is nth power).

Geometrically, on a smooth thing, this means you take the sum of all the Weil divisors as a homology class, and then take the Poincare dual class in $H^2(X;\mathbb{Z})$.

I have a silly answer to your second question. Take a disjoint union of an elliptic curve with any other curve, and set L to be a nontrivial degree 0 bundle on the elliptic curve and trivial on the other curve.