$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ 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). $