Languages that are not subset, but are union
Yes. Let $L_1=\{a\}, L_2=\{\epsilon\}$. Obviously both languages are regular and neither is a subset of the other. We then have 
$$
(L_1\cup L_2)^*=(a+\epsilon)^*=a^*
$$ 
and since 
$$
L_1^*=a^*,\quad L_2^*=\epsilon^*=\epsilon 
$$
we have
$$
L_1^*\cup L_2^*=a^*\cup \epsilon = a^*
$$