= Solution
Use the <bounded-run deterministic finite automaton> with states
$$
q_0,q_1,\ldots,q_7,q_{\rm dead}.
$$
State $q_j$ records that the current suffix consists of exactly $j$ consecutive zeros. Every $q_0,\ldots,q_7$ is accepting; $q_{\rm dead}$ is rejecting. Reading $1$ sends every accepting state to $q_0$, while reading $0$ sends $q_j$ to $q_{j+1}$ for $j<7$ and sends $q_7$ to $q_{\rm dead}$. The dead state loops on both symbols. This finite automaton accepts exactly the words with no run of eight zeros, so
$$
\boxed{\text{the language is regular}.}
$$
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