Sum of 2^Pi mod 1000000007 for all i where Pi is sum of numbers in ith subset of a set X
If the set you're using is $X=\{x_1,x_2, \dotsc,x_n\}$, then the expression you want to evaluate is equivalent to
$$
(2^{x_1}+1)(2^{x_2}+1)\dotsm(2^{x_n}+1)\bmod{1000000007}
$$
This isn't too hard to prove; here's an example to help you see what's going on:

Suppose $X=\{a,b,c\}$. Then the subsets of $X$ are, obviously,
$$
\{\},\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}
$$
and the corresponding sums of elements in those subsets will be
$$
0,a,b,c,a+b,a+c,b+c,a+b+c
$$
leading to the sum
$$
2^0+2^a+2^b+2^c+2^{a+b}+2^{a+c}+2^{b+c}+2^{a+b+c}=(2^a+1)(2^b+1)(2^c+1)
$$
The general result can be established by a fairly simple induction proof.