Let $X_1$ and $X_2$ be random variables with CDF $F_X$. Let $Y$ be a random variable with PDF $f_Y$. Let $X_1\perp Y$ and $X_2\perp Y$.
I know that $$P(X_1<Y,X_2<Y)=\int_{-\infty}^{\infty}P(X_1<y,X_2<y|Y=y)f_Y(y)dy\stackrel{\perp}{=}\int_{-\infty}^{\infty}P(X_1<y)P(X_2<y)f_Y(y)dy=\int_{-\infty}^{\infty}[F_X(y)]^2f_Y(y)dy.$$
But how to find $P(X_1>Y,X_2>Y)$? Will it be $1-P(X_1<Y,X_2<Y)$? Or $(1-P(X_1<Y))(1-P(X_2<Y))$? Or something completely different?
Assuming the probability mass for $X_1=Y$ is zero, we have that $X_1>Y$ is equivalent to $\neg (X_1<Y)$. Distributing a negation over a conjunction gives a disjunction: $$1-P((X_1<Y)\land (X_2<Y))=$$ $$P( \neg ((X_1<Y)\land (X_2<Y))) =$$ $$ P( \neg (X_1<Y)\vee \neg(X_2<Y))=$$ $$ P( (X_1>Y)\vee (X_2>Y))$$. But you want the probability $P( (X_1>Y)\land (X_2>Y))$. So your first guess doesn't work.
Your second guess doesn't work because by finding the probability of the two conditions separately and multiplying them, you're treating them as independent. But clearly they are dependent; if $X_1>Y$, then it's more likely that $Y$ is small, which makes it more likely that $X_2>Y$.
To find the probability, just reverse the inequalities in the integrals. For instance, instead of $\int_{-\infty}^{\infty}P(X_1<y,X_2<y|Y=y)f_Y(y)dy$, take $\int_{-\infty}^{\infty}P(X_1>y,X_2>y|Y=y)f_Y(y)dy$