How can I calculate $P(2-XSuppose $X,Y$ are independent $N(1,1)$. 
How can I calculate $P(2-X<Y<X)$?
 A: It will help to draw pictures of the regions involved as you go through this. 
[Indeed, you could even start by drawing up an $X,Y$ plane with the bivariate mean and some contours drawn in (e.g. at radius 1 and 2). Draw on the region in the $X,Y$ plane that you want the probability for; this should already show you how to proceed. Or you can skip that and move straight to the next step.]
Consider $X_1=X-1$ and $Y_1=Y-1$. Write the inequalities in terms of $X_1$ and $Y_1$ and look at what region that defines in a diagram of the $X_1,Y_1$ plane.
You may be able to answer by inspection of the relevant region in the $X_1,Y_1$ plane (it's quite plain what the answer is, and that answer can be formalized by a simple appeal to symmetry).
Failing that, if you can see (or show) that  $U=X_1-Y_1$ and $V=X_1+Y_1$ are also bivariate normal and uncorrelated (even simply on the basis that $kU,kV$ for a particular choice of $k$ is a rotation about the origin of a circular symmetric bivariate distribution, for example), then consider the effect on the inequalities to reparameterizing to $U,V$ (that is, write the inequalities in terms of $U$ and $V$).
If you look at the resulting region in the ($U,V$) plane you should be able to see how to write a somewhat more formal argument that establishes the answer quite easily.
A: So if you had two real values $a,b$ you would compute the probability $P(a<Y<b)$ as $$P(a<Y<b)=\int_a^b f_Y(y)dy,$$ right? ($f_Y(y)$ is the density of $Y$).
Now, since the boundaries are also random, you have to make this calculation integrating for every possible value of the boundaries, i.e.,
$$P(2-X<Y<X)=\int_{-\infty}^{\infty}\left(\int_{2-x}^x f_Y(y)dy\right)f_X(x)dx.$$
