Hardness of exact binomial tail bounds
Theorem Exactly computing $F(w,t)$ is #P-hard.

proof.  Reduce from the counting version of Subset Sum, the #Subset Sum problem, defined as follows. Input is a set $W = w_1, \ldots, w_n$ of non-negative integers, and an integer $t$. The task is to count the number of subsets $W' \subseteq W$ that add up to exactly $t$. This problem is known to be #P-hard, even when for all $i$, $w_i \leq t$ and $t = 2^{O(n)}$. In particular, the $w_i$'s are $O(n)$ bit integers.  

Suppose now we could compute F in polynomial time. Then we could solve the #Subset Sum problem by outputting:

$$2^n \cdot \left( F(\frac{W}{t}, 1-\frac{1}{t}) - F(\frac{W}{t}, 1) \right) = 2^n \cdot P\left[\sum_{i=1}^n \frac{w_iX_i}{t} = 1 \right],$$
which is precisely the number of subsets of $W$ that add up to $t$.