Regular Expressions with at least 2 0's and at most 1
Your answer is correct. You can obtain a slightly shorter expression by deleting two of the $\epsilon$ in your expression. Indeed
\begin{align}
(1+ \epsilon)000^* + 0^*0(1+\epsilon)00^* + 000^*(1+\epsilon) &amp;= (1+ \epsilon)000^* + 0^*0100^* + 000^*1 \\
&amp;= 1000^* + 0^*0(1+\epsilon)00^* + 000^*1\\
&amp;=1000^* + 0^*0100^* + 000^*(1+\epsilon)
\end{align}
Finally, if you want to avoid $\epsilon$, you could use one of the expressions
$$
000^* + 1000^* + 0^*0100^* + 000^*1 \quad \text{or} \quad 0^*(00 + 100 + 010 + 001)0^*
$$
which might be easier to read.