NPC problem with sets
I will prove it is NP-hard leaving the proof that it is in NP to you.

Let $\phi = c_1\land c_2\dots\land c_m$ be a 3SAT formula. Define 
$$S = \{l \mid l \text { is a literal of  } \phi \}$$
$$S_i = \{l \mid l \text { is a literal of } c_i\}$$

Since each clause has to have at least one literal with TRUE value we have the following constraints
$$l_i=1 \leq |T \cap S_i|\leq 3 = h_i$$ 
Finally, we have to make sure that $T$ contains no variable $x$ (of $\phi$) and its negation at the same time
$$S_{x} = \{x,\overline{x} \} $$
$$l_x=0\leq |T \cap S_x| \leq 1=h_x$$

So, if $\phi$ indeed has a solution then $T$ consists of those literals whose true-value is TRUE. Then it is trivial that all constraints satisfy since $T$ may not have a variable and its negation at the same time and each clause has at least one TRUE literal. 

On the other hand, if we have a set $T$ satisfying the above-described constraints then $\phi$ can be constructed by ORing each literal in $S_i$ (by ORing with arbitrary variables if $|S_i| &lt; 3$) and ANDing all clauses corresponding to $S_i$. This formula is satisfiable since may make each clause TRUE by making true each literal in $S_i$ and since not $x$ and $\overline{x}$ are in $T$.