Answer:
(a) AABBCC × aabbcc->AaBbCc     → 1
(b) AABbCc × AaBbCc->AAbbCC   → 1/32 
(c) AaBbCc × AaBbCc->AaBbCc     → 1/8 
(d) aaBbCC × AABbcc->AaBbCc    → 1/2
Explanation:
In such cases we calculate the probability of each allelic pair separately. 
(a) AABBCC × aabbcc -> AaBbCc  
Aa  Aa   Aa  Aa           Bb  Bb  Bb  Bb         Cc  Cc  Cc  Cc
↓                                   ↓                                ↓  
4/4 = 1                          4/4 = 1                       4/4 = 1     
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc   = 1 x 1 x 1 = 1.                 
(b) AABbCc × AaBbCc -> AA bb CC
AA  AA   Aa  Aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc
↓                                                        ↓           ↓  
2/4                                                    1/4          1/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.
In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.
So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.
(c) AaBbCc × AaBbCc -> AaBbCc
AA  Aa   Aa  aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc
         ↓                                  ↓                               ↓  
        2/4                               2/4                           2/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.
So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.
(d) aaBbCC × AABbcc->AaBbCc
Aa  Aa   Aa  Aa               BB  Bb  Bb  bb          Cc  Cc  Cc  Cc
       ↓                                        ↓                                ↓  
4/4 = 1                                       2/4                          4/4 = 1     
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.