Independence of sigma algebra I am trying to establish whether the following is true (my intuition tells me it is), more importantly if it is true, I need to establish a proof.  
If $X_1, X_2$ and $X_3$ are pairwise independent random variables, then if $Y=X_2+X_3$, is $X_1$ independent to $Y$? (One can think of an example where the $X_i$ s are Bernoulli random variables, then the answer is yes, in the general case I have no idea how to prove it.)
A related problem is:
If $G_1,G_2$ and $G_3$ are pairwise independent sigma algebras, then is $G_1$ independent to the sigma algebra generated by $G_2$ and $G_3$ (which contains all the subsets of both, but has additional sets such as intersection of a set from $G_2$ and a set from $G_3$).
This came about as I tried to solve the following:
Suppose a Brownian motion $\{W_t\}$ is adapted to filtration $\{F_s\}$, if $0<s<t_1<t_2<t_3<\infty$, then show $a_1(W_{t_2}-W_{t_1})+a_2(W_{t_3}-W_{t_2})$ is independent of $F_s$ where $a_1,a_2$ are constants.
By definition individual future increments are independent of $F_s$, for the life of me I don't know how to prove linear combination of future increments are independent of $F_s$, intuitive of course it make sense...
Any help is greatly appreciated.

Thank you for the hints, I think I have made progress in my understanding, please confirm these if possible.
In one of the assumptions of the Brownian motion $\{W_t\}$ is adapted to a filtration $F_t$, do we assume:
A) Each increment $(W_t-W_s)$ is independent of $F_u$ for $0<u<s<t$. 
OR
B) Any number of disjoint increments (in the future of time $s$) and $F_s$ are mutually independent.
Under B) then the assumption answers my problem.  Under A), if the filtration 
is the one generated by the Brownation motion, then the required mutual independence can be deduced from the mutual independence of the Brownian 
increments.  However, under A) if the filtration is not necessarily the one generated by the Brownian motion, is it still possible to prove
the required mutual independence?  If so please help, I spent a longggg time trying to work it out. 
Many thanks.
 A: No, $X_1$ and $Y$ need not be independent.  Consider a standard example of three pairwise independent random variables that are not independent: $X_2$ and $X_3$ independent, each having values $-1$ and $1$ with probabilities $1/2$, $X_1 = X_2 X_3$.  Then $Y = X_2 + X_3$ is not independent of $X_1$, in fact $Y = 0$ if and only if $X_1 = -1$.
A: Robert Israel has already given a counterexample to your first question, and it also gives a counterexample for your second question (let $G_i = \sigma(X_i)$).
As for your question about Brownian motion, I think it isn't worded quite right: to say $W_t$ is adapted to $\{F_t\}$ usually just means $W_t \in F_t$ for each $t$.  You also need to know that $W_t - W_s$ is independent of $F_s$, which doesn't follow from adaptedness alone.  This stronger assumption is sometimes stated as "$W_t$ is a Brownian motion with respect to the filtration ${F_t}$".
Under this assumption, the answer to your question is yes, and the key is that $F_s$, $W_{t_2} - W_{t_1}$ and $W_{t_3} - W_{t_2}$ are not merely pairwise independent, they are mutually independent.  I recommend trying to prove this.  Once it is done, it is not too hard to get a proof for your problem.  Give it a try, and post back if you have further questions.
Edit: As to your revised questions (presently posted as another answer), the usual assumption is (A), but (A) implies (B).  To see why, suppose $s \le t_1 \le t_2 \le t_3 \le t_4$, $A \in F_s$, $B \in \sigma(W_{t_2} - W_{t_1})$, $C \in \sigma(W_{t_4} - W_{t_3})$.  We have to show $$P(A \cap B \cap C) = P(A) P(B) P(C).$$
The key is that $A \cap B \cap C = (A \cap B) \cap C$, and $A \cap B \in F_{t_2}$ (verify this).  But $C$ is independent of $F_{t_2}$.  Now I expect you can finish the argument, and extend it to any number of increments.
