Is it possible that the subtraction between two undecidable languages is regular? If $L_1$ and $L_2$ are both non-decidable languages (Not decidable, so can be SD 
or $\lnot$SD), is it possible that $L_1-L_2$ is regular and $L_1-L_2\neq\phi$, where $\phi$ is the empty set?

What's a good way of tackling this question? Me and some classmates have been stuck on this question for a while, and this was the best we've got:

It should be possible given the following example:

Let $L_1=H\cup a^*$ and $L_2=H$, where $H=\{&lt;M,w&gt;:$ Turing machine $M$ halts on $w$$\}$, so $L_1-L_2=L_1\cap \lnot L_2=(H\cup a^*)\cap \lnot H=\lnot H\cap a^* = a^*$ which is regular. (we are saying $H$ does not contain $a^*$, so $\lnot H$ must contain $a^*$, hence why the intersection gives $a^*$.)

But we are not confident this is correct due to the last part with the intersection does not feel right. What's the answer to this question?
Your example almost works. You need to make sure that the regular part is disjoint from the non-decidable part.

Suppose our alphabet has at least two symbols, say $a$ and $b$. Consider any undecidable langauge $H$, for example the halting set, and define
$$L_2 = \lbrace b w \mid w \in H\rbrace$$
and
$$L_1 = L_2 \cup \lbrace a \rbrace.$$
Now it is obvious that $a \not\in L_2$ because all words in $L_2$ start with $b$. Also, $L_2$ and $L_1$ are undecidable because there are easy reductions from $H$ to them, and back. And of course we have
$$L_1 \setminus L_2 = \{a\},$$
which is decidable. You can make the example more complicated by taking any decidable langauge $D$ and define $L_2$ as before, but change $L_1$ to
$$L_1 = L_2 \cup \{a w \mid w \in D\}.$$
This way $L_1 \setminus L_2$ is computably isomorphic to $D$.