Have a roll (1)!

This can be very simply done manually.

First, take any of the 6 digits (1,1,3,5,8,13) on one dice with other dice as 2 (6 cases)

Since order of rolling can be different so we will take twice [12 cases]

Now do similarly with one dice as 8 and other as any of the 5 digits (1,1,3,5,13) as 2 & 8 has already been counted above (5 cases)

Again twice will make [10 cases]

Total favourable cases are 12+10=22.

Total possible cases are 7x7=49

So probability=22/49=a/b

a+b=71

The answer is $\frac { 22 }{ 49 }$

Since the number has to be even, the two dice must have at least a $2$ or a $8$ on them.

Since one number has to be greater than the other, the $2$ numbers cannot be the same.

The probability of the two dice rolling a $2$ or a $8$ is:

$2 \times \frac { 2 }{ 7 } =\frac { 4 }{ 7 }$

$\times2$ because the order of rolling can be different.

Given that the $2$ numbers cannot be the same, the probability of rolling double $2$ or $8$ is (Taking into consideration $2$ and $8$ as I am going to minus this from the value above):

$2\left[ \frac { 2 }{ 7 } \times \frac { 1 }{ 7 } \right] =\frac { 4 }{ 49 }$

Minus this from the value at the top gives: $\frac { 4 }{ 7 } -\frac { 4 }{ 49 } =\frac { 24 }{ 49 }$

Note that by calculating the values above, I had over calculated by adding the sets that satisfy both conditions for both of its dice's value: $2,8$ and $8,2$

Therefore, the answer is:

$\frac { 24 }{ 49 } -\frac { 2 }{ 49 } =\frac { 22 }{ 49 }$ , making a+b=22+49=\boxed { 71 }

Firstly, for one number to be greater than the other, the results from the two rolls cannot be the same number. Taking into account that there are two '1's on the dice, the probability of the dice having the same number for one roll is $\frac{1}{7}$ ,

and thus fulfilling the criteria be $\frac{6}{7}$ .

Next, for the product of the two results to be even, at least one of the rolls must give an even number, ie 2 or 8.

The probability that one roll will give an even number is thus $\frac{2}{7}$ per roll.

Hence, each roll will give a $\frac{6}{7}\times\frac{2}{7}=$ $\frac{12}{49}$ chance of fulfilling both criteria,

and two rolls hence $\frac{12}{49}\times2=\frac{24}{49}$ .

However, this result would include the repeated inclusion of one set of {2,8} and {8,2} possibility.

Thus, $\frac{24}{49}-\frac{2}{49}=\frac{22}{49}=\frac{a}{b}$ .

Therefore, $a+b=22+49=\boxed{71}$ .