No. For example, take four independent identically distributed random variables $X_1,X_2,Y_1,Y_2$ and then swap $Y_1,Y_2$ if necessary so that they are in the same order as $X_1,X_2$.
For the first one:
Notice first that $$N=n \iff (X_1 =0\cup Y_1=0)\cap(X_2 =0\cup Y_2=0) \cdots \cap(X_{n+1} =1\cap Y_{n+1}=1) $$
Let's abuse notation to abbreviate $p(x_i) = P(X_i = x_i)$ , $p(n)=P(N=n)$, etc. Then
$$p(x_i \mid n)=\frac{p(n\mid x_i) p(x_i)}{p(n)} \tag1$$
But, for $1\le i\le n$:
$$
\begin{align}
p(n)&=(1-p^2)^n p^2 = q^{n}(1+p)^{n} p^2 \tag 2 \\
p(n\mid x_i)&=(1-p^2)^{n-1} p^2 q^{x_i} \tag 3 \\
p(x_i)&= p^{x_i} q^{1-x_i} = q \, (p/q)^{x_i} \tag 4
\end{align}
$$
Hence
$$p(x_i\mid n)=\frac{q^{x_i} q \, (p/q)^{x_i}}{1-p^2}=p^{x_i} (1+p)^{-1}
\tag 5$$
and
$$\prod p(x_i \mid n) = p^{w_x} (1+p)^{-n} \tag 6$$
where $w_x=\sum x_i$ is the weight of the bistring ${\bf x}=(x_1 \cdots x_n)$
Now, we compute the joint prob:
$$p({\bf x}\mid n)=\frac{p(n\mid {\bf x}) p({\bf x})}{p(n)}$$
with
$$p({\bf x})=p^{w_x} \, q^{n-w_x} \tag 7$$
$$p(n\mid {\bf x})=q^{w_x } p^2 \tag 8$$
So
$$p({\bf x}\mid n) = \frac{p^{w_x} q^n p^2 }{q^{n}(1+p)^{n} p^2}=p^{w_x} (1+p)^{-n} \tag 9$$
which coincides with $(6)$.
For the second part, I get, from $(5)$
$$p(X_i=1 \mid n) = \frac{p}{1+p}=\frac{pq}{1-p^2}$$
which does not coincide with your value. It coincides, however with
$$ P(X_i=1 \mid X_i Y_i=0) = \frac{ P(X_i Y_i=0\mid X_i=1 ) P(X_i=1)}{P(X_i Y_i=0)}=\frac{q \, p}{1-p^2} $$
Care to check that?
Best Answer
If they were independent we would have $Ex_1^{2}y_1^{2}x_2^{2}y_2^{2}=Ez_1^{2}z_2^{2}=Ez_1^{2}Ez_2^{2}=Ex_1^{2}y_1^{2}Ex_2^{2}y_2^{2}$. Consider the case where $y_1=y_2$. We get $Ex_1^{2}y_1^{4}x_2^{2}=(Ex_1^{2}y_1^{2})^{2}$ since $x_1y_1$ and $x_2y_1$ have the same distribution. Thus $Ey_1^{4}=((Ey_1)^{2})^{2}$ which implies that $y_1^{2}$ is a constant (by conditon for equality in Cauchy-Schwarz inequality). This contradiction shows that the result is not true when $y_1=y_2$.
If $y_i$'s are also independent then the result is true.