The angle 495° is more than one complete revolution counter-clockwise
from the right side of the x-axis.  So we subtract 1 revolution or
360° from 495° getting 135°.  So we can forget the first revolution.
We now have just this:



Next we see that 135° is in the second quadrant. Let's indicate the 
reference angle in red, which is the smallest possible amout of 
rotation to get to the x-axis from the terminal side:



We subtract 135° from 180° to get 45°, which is the reference angle
indicated by the red arc.



Now we remember our 45°-45°-90° right triangle:



The cosecant is the hypotenuse over the opposite which
is  or just .  We remember that
angles in the second quadrant have positive sines and cosecants,
so the answer is
               _
csc(495°) = +Ö2

Edwin