Problem 1:

(I)     A  +  B  =  CD
        +     +      +
(II)   CE  +  F  =  CF
       ---------------
(III)  CA  + CC  =  DE
      (IV)   (V)   (VI) 


Problem 2:

(I)     A  +  B  =  CD
        +     +      +
(II)   CE  +  F  =  CF
       ---------------
(III)  CG  + CC  =  HE
      (IV)   (V)   (VI) 



Solution 1:
(V)     C = 1
(II.1)  E = 0
(III.1) A = 9
(III)   D = 3
(I.1)   B = 4
(VI.1)  F = 7

        9  +  4  =  13
        +     +      +
       10  +  7  =  17
       ---------------
       19  + 11  =  30


Solution 2:
(V)     C = 1
(II.1)  E = 0
(IV.1)  A = G
(III.1) A = 9
(III)   H = 3
(I.1)   B = D + 1
(VI.1)  D + F = 10
So assuming that different letters may be the same digit 
we get solutions for D in {1,2,3,4,5,6,7,8}.


References: 
  J. Petigk, Mathematik in der Freizeit, 1998, p15