Here’s what he was thinking for (1). You know that the language $L_1$ is regular. If your language $L$ were regular, its complement $\overline{L}$ would also be regular. Moreover, the intersection of two regular languages is regular, so $L_1\cap\overline{L}$ would be regular. But $L_1\cap\overline{L}=L_2$.
Proof: Clearly $L_2\subseteq L_1\cap\overline{L}$. On the other hand, if $w\in L_1\cap\overline{L}$, then $w\in L_1$, so $w=0^n1^m$ for some $n,m\ge 0$. Moreover, $w\in\overline{L}$, so $w\notin L$, and therefore $n=m$. $\dashv$
Thus, $L_2$ would have to be regular. But you almost certainly proved in class or in the text that $L_2$ is not regular: it’s just about the simplest example of a context-free language that isn’t regular.
For (2) I assume that you meant the language to be $L=\{0^n1^m:n\ge 2m\ge 0\}$. Let $p$ be the pumping length, and let $w=0^{2p}1^p$. The pumping lemma guarantees that we can write $w$ as $xyz$, where $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for all $k\ge 0$. Then $xy$ is contained in the first $p$ letters of $w$, so $y=0^i$ for some $i>0$, and $xy^0z=0^{2p-i}1^p\notin L$, since $2p-i\not\ge 2p$. Remember, you can also pump down to $k=0$; people often forget this.
For (3) and (4), $w00$ and $00w$ do mean the concatenation of $w$ and $00$ (in the two possible orders), but I don’t see how you got a union out of this. The union of two languages is the set of words that belong to at least one of the two languages; concatenation is a different animal altogether. Before I suggest an approach to these, I need to know whether you’ve already studied the connection between finite state automata and regular languages; if you have, that’s probably the easiest way to approach these two questions.
Added: For (3), you know that there is an NFA that recognizes $L$. I don’t know exactly what formalism you use for NFAs, so I’ll specify it as $M=\langle\Sigma,S,s_0,\delta,A\rangle$, where $\Sigma$ is the alphabet, $S$ is the state set, $s_0$ is the initial state, $\delta$ is the transition function, and $A$ is the set of acceptor states. Let $t$ and $u$ be two new states not in $S$, and let S'=S\cup\{t,u\}. $\Sigma$ and $A$ remain unchanged. The new initial state is $u$. The changes to $\delta$ are minor: $u$ has a $0$ transition to $t$, $t$ has a $0$ transition to $s_0$, and all of the original transitions remain unchanged. The resulting NFA clearly accepts $L_1$.
From what you said in the comment, I expect that you can probably do $L_2$ yourself on the basis of this model.
Added: The solution above for (3) is actually based on a misreading of the questions: the NFA that I produced recognizes $\{00w:w\in L\}$, not the $L_1$ of the problem. In other words, I was prepending $00$ to each word of $L$ instead of recognizing the words that appear as suffixes to $00$ in words in $L$. As has been pointed out several times in the comments, modifying a DFA that recognizes $L$ to recognize $L_1$ instead is very straightforward.
If the DFA recognizing $L$ is $M=\langle\Sigma,S,s_0,\delta,A\rangle$, just change the initial state from $s_0$ to $\delta(\delta(s_0,0),0)\;,$ the state reached by $M$ after two $0$ transitions from $s_0$, and you’ll have a DFA that recognizes $L_1$. To recognize $L_2$, just change the set $A$ of acceptor states to A'=\{s\in S:\delta(\delta(s,0),0)\in A\}\;, the set of states from which two $0$ transitions take you to an acceptor state of $M$.