Whether or not $X_1$,...,$X_n$ are independent and exchangeable For some n = 1,2,..., let $Y_1$,...,$Y_{n+1}$ denote iid real-valued random variables. 
Define $X_j$ = $Y_j$$Y_{j+1}$, $\hspace{10mm}$j=1,...,n
a) Are $X_1$,$X_2$,...,$X_n$ independent?
b) Are $X_1$,$X_2$,...,$X_n$ exchangeable?
Attempts: 
a) No. 
Counterexample: Let Y be a Bernoulli (0,1) distribution, with p = 1/2, where $Y_1$ = 0 and $Y_2$ = 1. 
Then $X_1$ = $Y_1$$Y_{2}$ 
Now $E(Y_1*Y_2)%$ = $E(0*1)$ = $E(0)$ = 1/2 
However, $E(Y_1)%$$E(Y_2)%$ = 1/2*1/2 = 1/4
Since $E(Y_1*Y_2)%$ $\neq$ $E(Y_1)%$$E(Y_2)%$, the $X_i$'s are not independent.
b) Not sure.
 A: Your reasoning on a) is not valid yet the good idea is there. Let's keep your example with the $X_{i}$'s being independant Bernouilli variables with $p=0.5$.
Then, $\mathbb{E} X_{1}X_{2} = \mathbb{E} Y_{1}Y_{2}^{2} Y_{3} = \mathbb{E}Y_{1} \mathbb{E}Y_{2}^{2} \mathbb{E}Y_{3}$ because of the independance of the $Y_{i}$s. 
Since $\mathbb{E}Y_{2}^{2} = p$, we have $\mathbb{E} X_{1}X_{2} = p^{3}$. 
On the other side, it is easy to see that $\mathbb{E}X_{i} = p^{2}$. 
If $X_{1}$ and $X_{2}$ were independant, we should have $ p^{3} = \mathbb{E} X_{1}X_{2} = \mathbb{E} X_{1} \mathbb{E}X_{2} = p^{4}$, which is not true.
What do you mean by "exchangeable variables ?" Here is my definition : $X_{1}, ... , X_{n}$ are exchangeable if, for every measurable function $f : \mathbb{R}^{n} \to \mathbb{R}_{+}$, and for every permutation $\sigma \in \mathfrak{S}_{n}$, we have : 
$$\mathbb{E}f(X_{\sigma(1)}, ..., X_{\sigma(n)}) = \mathbb{E}f(X_{1}, ..., X_{n}).$$
Let's try with $n=2$. We have $f(X_{1}, X_{2}) = f(Y_{1}Y_{2}, Y_{2}Y_{3}) = f \circ g (Y_{1}, Y_{2}, Y_{3})$, where we defined $g : \mathbb{R}^{3} \to \mathbb{R}^{2}$ by : 
$$g : (x,y,z) \to (xy,yz) $$
Then, $f \circ g$ is measurable and since $(Y_{1}, Y_{2}, Y_{3})$ is i.i.d., we have : 
$$\mathbb{E}f f(X_{1}, X_{2}) = \mathbb{E}f \circ g (Y_{1}, Y_{2}, Y_{3}) = \mathbb{E}f \circ g (Y_{3}, Y_{2}, Y_{1}) = \mathbb{E}f (Y_{3}Y_{2}, Y_{2}Y_{1}) = \mathbb{E}f(X_{2}, X_{1})$$
Therefore,  $X_{1}$ and $X_{2}$ are exchangeable. I'm not sure the same result holds with $n > 2$, but you should try to adapt the proof (which might reveal tricky).
edit. Okay, here is a counterexample with $n=3$.
Let's have $f(x,y,z) = xz$, nd $\sigma = (2,3,1)$. We have : 
$$\mathbb{E}f(X_{\sigma(1)}, X_{\sigma(2)}, X_{\sigma(3)}) = \mathbb{E}f(X_{2}, X_{3}, X_{1})  = \mathbb{E}Y_{2}Y_{3}Y_{1}Y_{2} = p^{3}$$
On the other side, we have : 
$$\mathbb{E}f(X_{1}, X_{2}, X_{3}) = \mathbb{E}Y_{1}Y_{2}Y_{3}Y_{4} = p^{4}$$
In conclusion, $X_{1}, X_{2}$ and $X_{3}$ are not exchangeable. The $n=2$ cas was deceptive !
