Solution: The sequences that meet the requirements have the following two possible forms:
(1) Starting with F: $F \cdots F Z \cdots Z F \cdots F Z \cdots Z$;
(2) Starting with Z: $\mathrm{Z} \cdots \mathrm{ZF} \cdots \mathrm{FZ} \cdots \mathrm{ZF} \cdots \mathrm{F}$.
Since the sequence is required to have exactly 3 $\mathrm{ZF}$, if the sequence belongs to (2), it should have the form
$$\text { Z } \cdots \underline{Z F} \cdots F Z \cdots Z F \cdots F Z \cdots Z F \cdots F \text {. }$$

which contains only 2 $\mathrm{FZ}$, failing to meet the requirement of 4 $\mathrm{FZ}$. Therefore, the sequence that meets the requirements can only be of form (1).

Since the sequence has exactly 4 $\mathrm{FZ}$, when considering the sequence with exactly two ZZ, it can be divided into the following two categories: $\qquad$ ZZZ $\qquad$
$\qquad$
$\qquad$ Z, $\qquad$ ZZ $\qquad$ ZZ $\qquad$ Z $\qquad$ Z
and the different positions of $\mathrm{Z}$, with the spaces filled with $\mathrm{F}$.
Let the number of $\mathrm{F}$ in each space be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, then $x_{1}+x_{2}+x_{3}+x_{4}=9$ has $\mathrm{C}_{8}^{3}=56$ positive integer solutions.

On the other hand, for (1), the position of ZZZ has 4 possibilities, and for (2), the arrangement of $\mathrm{ZZ}, \mathrm{ZZ}, \mathrm{Z}, \mathrm{Z}$ has 6 possibilities, so the arrangement of $\mathrm{Z}$ has 10 possibilities.
Therefore, the number of sequences that meet the requirements is
$$10 \times 56=560 \text { (sequences) }$$