Regular Expressions with at least 2 0's and at most 1 I think the answer should be:

$(1+ \epsilon)000^* + 0^*0(1+\epsilon)00^* + 000^*(1+\epsilon)$

But I am not sure if this is the right answer. Can someone explain the correct answer?
And if it is correct how can I shorten this up?

Also what is the nfa for the regular expression:
∅*

i have one more problem:  i can't surely understand the meaning of Σ* symbol. As far as I understand it should be (0+1)* if Σ = (0, 1) and (a+b+c)* if 
Σ = (a,b,c) and so on. Please someone clarify this.
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.