$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ But, The normal distribution is symmetric so no calculation is needed: The tuple is equally likely to take a value in any of the 8 areas shown below.

       Y  ^     / *****
          |    / ******
          |   / *******
          |  / Y = X **
          | / *********
          |/ **********
  ------------------>
         /|         X
        / |
       /  |
          |

For non-believers:

$P(X>0) = \int_0^\infty e^{-x^2/2} dx = \frac{1}{2\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-x^2/2} = \frac{1}{2}.$ $P(X>Y\mid X,Y>0) = \frac{2}{2\pi}\int_0^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{1}{\pi}\int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr\\ =\frac{1}{4}\int_0^{\infty} r e^{-0.5 r} dr = \frac{1}{4}(1-0). $

$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ But, The normal distribution is symmetric so no calculation is needed: The tuple is equally likely to take a value in any of the 8 areas shown below.

       Y  ^     / *****
          |    / ******
          |   / *******
          |  / Y = X **
          | / *********
          |/ **********
  ------------------>
         /|         X
        / |
       /  |
          |

For non-believers:

$P(X>0) = \int_0^\infty e^{-x^2/2} dx = \frac{1}{2\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-x^2/2} = \frac{1}{2}.$ $P(X>Y\mid X,Y>0) = \frac{2\cdot 2}{(\sqrt{2\pi})^2}\int_0^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{2}{\pi}\int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr\\ =\frac{1}{2}\int_0^{\infty} r e^{-0.5 r} dr = \frac{1}{2}(1-0). $

Edit The original formulation asks for $P(X>Y\mid Y>0)$: $\frac{2}{(\sqrt{2\pi})^2}\int_{-\infty}^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{1}{\pi}(0 + \int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr)=\frac{1}{4}.$

$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ But, The normal distribution is symmetric so no calculation is needed: The tuple is equally likely to take a value in any of the 8 areas shown below.

       Y  ^*****/
          |****/
          |***/
          |**/ Y = X
          |*/
          |/
  ------------------>
         /|         X
        / |
       /  |
          |

For non-believers:

$P(X>0) = \int_0^\infty e^{-x^2/2} dx = \frac{1}{2\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-x^2/2} = \frac{1}{2}.$ $P(X>Y\mid X,Y>0) = \frac{2}{2\pi}\int_0^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{1}{\pi}\int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr\\ =\frac{1}{4}\int_0^{\infty} r e^{-0.5 r} dr = \frac{1}{4}(1-0). $

$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ But, The normal distribution is symmetric so no calculation is needed: The tuple is equally likely to take a value in any of the 8 areas shown below.

       Y  ^     / *****
          |    / ******
          |   / *******
          |  / Y = X **
          | / *********
          |/ **********
  ------------------>
         /|         X
        / |
       /  |
          | 

For non-believers:

$P(X>0) = \int_0^\infty e^{-x^2/2} dx = \frac{1}{2\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-x^2/2} = \frac{1}{2}.$ $P(X>Y\mid X,Y>0) = \frac{2}{2\pi}\int_0^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{1}{\pi}\int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr\\ =\frac{1}{4}\int_0^{\infty} r e^{-0.5 r} dr = \frac{1}{4}(1-0). $