It is certainly possible that $1 < [L:F] < \infty$, i.e. that the extension $F^{\mathrm{un}}/F$ be finite and non-trivial. The simplest example of this is $F = \mathbb{Q}(\sqrt{-5})$. Its Hilbert class field is $F(\sqrt{-1}) = \mathbb{Q}(\sqrt{-1},\sqrt{-5})$, and it can be shown that this field has a smaller root discriminant than any number field of degree at least eight, and hence that it has no unramified extensions. Therefore $F^{\mathrm{un}} = F(\sqrt{-1})$ in this case, and $[L:F] = 2$.

Continuing this, note that if $B(2d)$ is a lower bound on the root discriminants of totally imaginary number fields of degrees $\geq 2d$, and if $F$ is an imaginary quadratic field with root discriminant $< B(2d)$, then $[F^{\mathrm{un}}:F]$ is finite and $< d$. Using the available lower bounds on the $B(2d)$ (Odlyzko; Diaz), Yamamura [*Maximal unramified extensions of imaginary quadratic numer fields of small conductors*, Proc. Japan Acad. 1997] has verified in this way that $[F^{\mathrm{un}}:F] < \infty$ for all imaginary quadratic fields of conductors $\leq 420$. In fact, for these fields the Hilbert class field tower terminates at most at the third step $K_3$, and $K^{\mathrm{un}} = K_3$. Thus, if you exclude the nine fields of class number one, all imaginary quadratic fields of conductor $\leq 420$ have $1 < [F^{\mathrm{un}}:F] < \infty$.

Regarding your second question, in general it is not easy to determine $[F^{\mathrm{un}}:F]$; in particular, it is a difficult open problem to decide whether there are infinitely many $F$ with no unramified extensions. On the other hand, Golod-Shafarevich yields criteria for the Hilbert class field tower to be infinite and hence for $[F^{\mathrm{un}}:F] = \infty$; this is so for instance for all quadratic fields whose conductor has at least eight prime factors.