There is a standard technique for problems like this, where the language can be specified in the form
All words over some alphabet $\Sigma$ where [some condition] is satisfied by all contiguous substrings of a fixed length, $k$.
The key idea here is to define a collection of states $s_i$ where $0\le i \le |\Sigma|^k$. Basically, you are using the states to "remember" the most recently-seen $k$ characters. These states will be part of the finite automaton we'll build for our language.
In what follows, I'll use a simpler language than yours, so that the FA will be sufficiently small to fit here and still be readable. Let
$$
L=\{w\in\{0, 1\}^*\mid \text{every length-3 substring of $w$ contains at least 2 zeros}\}
$$
We'll define states $s_0, s_1, \dotsc, s_7$ to represent the three most recently-seen characters. It's convenient to do this representation in lexicographic order, so $s_0=\mathtt{000}$, $s_1=\mathtt{001}$, $s_2=\mathtt{010}$, $s_3=\mathtt{011}$, $s_4=\mathtt{100}$, $s_5=\mathtt{101}$, $s_6=\mathtt{110}$, and $s_7=\mathtt{111}$. In other words, $s_N$ will correspond to the 3-bit binary representation of $N$.
Now we'll add some more states to get from the start state to the $s$'s. All it takes is to construct a complete tree (binary in this example) having the $s$ states as leaves, like this:
All of the states in the FA above will be accepting states, except for the darkened ones. The $p_i$ states are all accepting, since we haven't yet seen a length-3 string, so we haven't violated the condition that defines the language $L$ and only the patterns $\mathtt{011}, \mathtt{101}$, $\mathtt{110}$, and $\mathtt{111}$ violate the condition of the language. These correspond to states $s_3, s_5, s_6, s_7$. Once we've entered one of these states, we reject the input word, so we might as well merge these into a single "dead" state, $d$, from which there will be no exit.
We're almost done; all that remains is to fill in the transitions among the $s_i$. Because of the way we chose the representations of the states, that's an easy task. If we're in state $s_i$, having just seen characters $b_1b_2b_3$, on input $b$ we'll pass to the state $s_j$, corresponding to the new pattern $b_2b_3b$. If you think about it for a moment, you'll see that $\delta(s_i, 0) = s_j$ where $j\equiv 2i\pmod8$ and $\delta(s_i, 1) = s_j$ where $j\equiv 2i+1\pmod8$. Thus, we'll have the following $s$ transitions (recall that we merged states $s_3, s_5, s_6, s_7$ into a single dead state, $d$):
$$\begin{array}{c|cc}
& 0 & 1 \\ \hline
s_0 & s_0 & s_1 \\
s_1 & s_2 & d \\
s_2 & s_4 & d \\
s_4 & s_0 & s_1 \\
d & d & d
\end{array}$$
completing the construction. The nice feature of this technique is that it's almost completely mechanical: the only hard part is making the $s$ transitions. The downside is that it produces a pretty big FA. Even in this simple example, we wound up with a 12-state FA, and if we had needed to look at the five most recent characters, we'd have a FA with 31 states just for the $p$s. In fact, if we apply the standard technique for DFA minimization, we would find that the minimal-state DFA for this example language would require only 7 states, with 6 of them being final.
