The given NFA \(M\) has five states (\(1, 2, 3, 4, 5\)) and includes \(\epsilon\)-transitions. To determine the language accepted by \(M\), consider the paths and transitions:
1. From state \(1\) to \(2\): The \(\epsilon\)-transition allows state \(2\) to be reached directly without consuming any input.
2. From state \(2\) to \(3\): Input \(0\) can be consumed repeatedly (loop on \(2\)) or proceed to \(3\) with a single \(0\). This corresponds to \((00)^*\).
3. From state \(1\) to \(4\): The \(\epsilon\)-transition allows state \(4\) to be reached directly without consuming any input.
4. From state \(4\) to \(5\): Input \(1\) can be consumed repeatedly (loop on \(5\)) after consuming \(1\). This corresponds to \((11)^*\).
5. Combining paths: The overall language includes all possible combinations:
- \(0^*\), representing sequences of \(0\)'s from state \(2\).
- \((1 + 0(00)^*)(11)^*\), combining sequences starting with \(1\) or sequences starting with \(0\) followed by \((00)^*\), and ending with \((11)^*\).
The regular expression representing the language is:
\[
0^* + (1 + 0(00)^*)(11)^*.
\]
Final Answer:
\[
\boxed{\text{(B)}}
\]