Can you cast the spell?
Pyth, 55 53 52 50 bytes

FN*Fmsm?k}kG^Gvkcd\/ceKc-rz0`Hd\,#=sN)I!.-NhK1B)E0


Try it online: Demonstration or Test harness

Notice that the time and memory complexity is really bad. So the second example doesn't work. I allocates about 1.6 GB of Ram before it crashes on my machine. 

Explanation

The explanation is for the 53 solution. The only difference is, that the initial parsing happens in the middle instead of the beginning. 

Kc-rz0"{}"dFN*Fmsm?k}kG^Gvkcd\/ceKc-rz0`H\,#=sN)I!.-NhK1B)E0


So here's the initial parsing. 

Kc-rz0`Hd
   rz0     convert input() to lowercase
  -   `H   remove all curly brackets (`H = "{}")
 c      d  split at the space
K          assign to K


So the input "{W},{R},{R} {2/W},{W/B}" gets converted into ['w,r,r', '2/w,w/b']. 

m               ceK\,    map each cost d of the costs split by "," to:
 s                         the sum of
  m         cd\/           map each value k of cost split by "/" to:
    k                        k
   ? }kG                     if k in "abcdef...xyz" else
        ^Gvk                 Cartesian product with "abc...yz" of int(k) repeats


So what does this do? The cost input '2/w,w/b' gets converted into:

[['aa', 'ab', 'ac', ..., 'zx', 'zy', 'zz', 'w'], 'wb']


Every string in ['aa', 'ab', 'ac', ..., 'zx', 'zy', 'zz', 'w'] satisfies {2/W} and every char in 'wb' satisfies {w/b}. 

Now we generate the Cartesian product of these lists (or strings) and see, if any combination can be produced with the mana-pool. 

FN*F...              )      for N in Cartesian product of ...:
       #   )                   while 1:
        =sN                      N = sum(N)
                               this flattens N
            I!.-NhK            if not (subtract mana pool from N):
                   1             print 1 (True)
                    B            break
                      E      else:
                       0       print 0 (False)