for part (a),(b)
use x-1=u , y+1=v ... then u will get 2 integrations of the form e^(-x^2)dx which is a well known integration its result is as shown below from mathematica integrator
**broken link removed**
u can learn about erf(x) here:
https://mathworld.wolfram.com/Erf.html
... now let's solve
(a)Pr(X>2,Y<0) = ∫∫fX,Y(x,y)dxdy
x:2 to inf , y:-inf to 0
make the variable substitution and solve
(b)Pr(0<X<2,|Y+1|>2) =∫∫ fX,Y(x,y)dxdy
we set fX,Y(x,y)=fX(x).fY
(statistically independent)
we do integration on x from 0 to 2
and integration on y we divide it into sum of 2 integrations : 1st from (-inf) to -3 and 2nd from 1 to (inf)
(c) Pr(Y>X)=∫fy
dy .... y: x to (inf)
where fY
=∫fX,Y(x,y)dx ... x: (-inf) to (inf)
the final solution should be function in x