II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$

If $F_{a} 、 F_{b} 、 F_{c} 、 F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$.
From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$

which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} 、 F_{b} 、 F_{d-1} 、 F_{d}$.
If $b \leqslant d-3$, then
$$
\left(F_{a}+F_{b}\right)+F_{d-1} \leqslant F_{d-2}+F_{d-1}=F_{d},
$$

which is a contradiction.
Therefore, $b=d-2$, at this point,
$$
F_{a}+\left(F_{d-2}+F_{d-1}\right)=F_{a}+F_{d}>F_{d} \text {. }
$$

Thus, the side lengths of the quadrilateral are $F_{a} 、 F_{d-2} 、 F_{d-1} 、 F_{d}$.
Hence, $d-b=2$.