Suppose a pair of random variable is independent from another pair, does it mean that each random variable is independent from the other?

I agree that the answers to your four questions are all "YES" (provided the typo in question 3 is fixed).

However, in view of your motivation, I'd like to point out that the way to derive the factorisation equation is like this. First, in view of the independence between $(x_1, y_1)$ and $(x_2, y_2)$, we have $$ p(x_1, x_2, y_1, y_2) = p(x_1, y_1) p(x_2, y_2).$$

Next, it's generally the case that $p(x_1, x_2, y_1, y_2) = p(y_1, y_2 | x_1, x_2) p(x_1, x_2)$ and $p(x_1, y_1) = p(y_1 | x_1)p(x_1)$ and $p(x_2, y_2) = p(y_2 | x_2) p(x_2)$. So $$ p(y_1, y_2 | x_1, x_2) p(x_1, x_2) = p(y_1 | x_1)p(x_1) p(y_2 | x_2) p(x_2)$$

Using independence between $x_1$ and $x_2$ once more (this is like what you were asking for in your denominator), we have $p(x_1, x_2) = p(x_1)p(x_2)$, so $$ p(y_1, y_2 | x_1, x_2) p(x_1) p(x_2) = p(y_1 | x_1)p(x_1) p(y_2 | x_2) p(x_2).$$

Cancelling factors of $p(x_1)p(x_2)$ gives $$ p(y_1, y_2 | x_1, x_2) = p(y_1 | x_1) p(y_2 | x_2) ,$$ which is what you were after.

Edit: To address the comment below...

Saying that $(x_1, y_1)$ and $(x_2, y_2)$ are independent is to say that $$ p(x_1, x_2, y_1, y_2) = p(x_1, y_1) p(x_2, y_2).$$

Integrating both sides with respect to $y_1$ and $y_2$, we have $$ p(x_1, x_2) = iint dy_1 dy_2 p(x_1, x_2, y_1, y_2) = iint dy_1 dy_2 p(x_1, y_1) p(x_2, y_2) \ = int dy_1 p(x_1, y_1) times int dy_2 p(x_2, y_2) = p(x_1) p(x_2), $$ which is what you wanted to prove.

YES, YES, NO and YES. If two sigma algebras are independent then any sub-sigma algebras of these are also independent. This proves 1) 2) and 4).

For 3) take two independent (non-constant) random variables $U$ and $V$ and take $X_1=Y_1=U, X_2=Y_2=V$.