This one could drive you buggy .....

First, since each of the bugs has $3$ edges it can travel along there are $3^{8} = 6561$ possible patterns of motion for the $8$ bugs.

For the bugs not to meet, there are two possibilities: (i) the pattern of all their moves combined forms two closed loops on opposite sides of the cube, and (ii) the pattern of all their moves combined forms a loop around the cube that visits each corner exactly once.

In case (i), there are $3$ pairs of opposite sides, and each loop can go in one of two directions. This gives us $3*2*2 = 12$ distinct patterns of motion.

In case (ii), we first note that at each corner exactly two adjacent edges must be on the loop, one leading to the corner and one leading away. If two such edges share a side of the cube, a third edge of that side must also be part of the loop in order for the loop to go through the last corner of that side before exiting that side. This forms a $U$ -shape within the loop. (Note that the case where a $4$ -th edge of a side is also part of the pattern of motion results in case (i), which we've already dealt with.)

Now to count these loops, we start by focussing on one corner. Two of the three adjacent edges must be part of the loop, which can be chosen in $\binom{3}{2} = 3$ ways. These two edges share a side, so as per the above discussion one of the remaining two edges must also be on the loop; once this choice is made, the rest of the loop can only be completed in one way. So there are $3*2 = 6$ possible loop paths, each of which can be traversed in one of two directions, resulting in a total of $6*2 = 12$ distinct patterns of motion.

Thus we end up with a desired probability of $\dfrac{12 + 12}{6561} = \dfrac{8}{2187}$ . This means that $a = 8, b = 2187$ and $a + b = \boxed{2195}$ .

The best way to solve a combinatorics puzzle is to simulate it.

Label the vertices a, b, c, d, e, f, g and the edges 1 to 8 and create a directed graph like this.

Imgur

When a bug traverses an edge in the opposite direction, it's sign is taken to be negative. For example a path from D to C is -6

Now, let's make a list of all the 8-tuples which represent all possible ways the ants could go

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paths = []
for a in [1,2,3]:
     for b in [-3,4,5]:
             for c in [6,-4,7]:
                     for d in [-1,-6,8]:
                             for e in [-2,9,10]:
                                     for f in [-8,-10,11]:
                                             for g in [-11,-7,12]:
                                                     for h in [-9,-12,-5]:
                                                             paths.append([a,b,c,d,e,f,g,h])

Now, every valid path must not have two ants travelling in the same edge, i.e, must not contain a edge and it's negative. Also, it must not contain two edges that lead to the same vertex

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def good_path(x):
     for i in x:
             if -i in x: return False
     if not (-1 in x)^(-2 in x)^(-3 in x): return False
     if not (3 in x)^(-4 in x)^(-5 in x): return False
     if not (4 in x)^(-6 in x)^(-7 in x): return False
     if not (1 in x)^(6 in x)^(-8 in x): return False
     if not (2 in x)^(-9 in x)^(-10 in x): return False
     if not (10 in x)^(8 in x)^(-11 in x): return False
     if not (11 in x)^(7 in x)^(-12 in x): return False
     if not (9 in x)^(5 in x)^(12 in x): return False
     return True

We are ready!

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good_paths=filter(good_path,paths)
print len(good_paths),len(paths)

So, there are 24 valid possibilities out of all $3^8$ possibilities.

Thus required probability is $\boxed{\frac{24}{3^8}}$

Let 1,2,3,4,1',2',3',4' be the bugs. There are 12 different 8-length cycles describing bug movements (such as 1-2-3-4-4'-3'-2'-1' - read: bug 1 goes towards bug 2, bug 2 towards bug 3, etc, bug 1' goes towards bug 1) and 12 different pairs of cycles of length 4 (such as 1-2-3-4, 1'-2'-3'-4') which gives bugs 24 ways to cross edges without meeting. We can pick two opposite sides of the cube in 3 ways and rotate bugs on each side clockwise or counterclockwise, so there are $3*2*2=12$ (non ordered) pairs of 4-length cycles. It's not difficult to write down 12 8-length cycles if we draw our cube like a planar graph: Since each bug can choose between 3 adjacent vertices, there are $3^8$ permutations. Therefore, probability is $24/3^8 = 8/3^7$ and the answer is $8+3^7 = 2195$ .