(I)    ABB  :   C  =  DB
         :      *      :
(II)     E  *   A  =   C
       -----------------
(III)   FC  :  GA  =   B
       (IV)    (V)   (VI)


Solution:
First we see that none of A, B, C, E is 1.
(III.1)  A * B == C  (mod 5)  and  A!=0, B!=0
(IV.1)   C * E == B  (mod 5)  and  C!=0, E!=0
Now if A = 5 or B = 5 then C = 5. So A!=5, B!=5.
Now if C = 5 or E = 5 then B = 5. So C!=5, E!=5.
Therefore no A, B, C, E is 0 (mod 5).
(I.1)    B * C == B  (mod 5)  =>  C == 1  (mod 5).
So we get C = 6. And {A,E} = {2,3} from (II).
(V) shows that A must be even. So we have A=2 and E=3.
(V)     G=1
(IV.1)  B=8
(VI)    D=4
(III)   F=9


 288  :   6  =  48
   :      *      :
   3  :   2  =   6
  ----------------
  96  :  12  =   8



References:

Walter Lietzmann;
  Lustiges und Merkw\"urdiges von Zahlen und Formen,
  Verlag Vandenhoeck & Ruprecht,
  G\"ottingen 1955, 8. Auflage, ISBN 3-525-39112-9
  Chap II.5: Vergilbte Manuskripte (p128-131, 21 problems)
  Chap II.13: Anagramme, Kryptogramme, Geheimschriften und dergleichen
  (p191-192, 10 problems)