Question: [![enter image description here](https://i.stack.imgur.com/wK2Bz.png)](https://i.stack.imgur.com/wK2Bz.png)

Is this problem wrong? Because I think from ordinary die space we can get results like(5,5), but we can never get results like this from the given space in this problem.

Answer: 
As the comments have said, we already have an approach for solving the problem. Let me be a little more elaborate.

Let the numbers be $a,b,c,d,e,f$. 

1) First of all, $12$ can only come on one occasion on an ordinary dice $(6+6)$, and no number greater than 12 appears. That tells you that $f=8$, so that $4+8=12$ is the only possibility.

2) Similarly, for $2$ coming only on one occasion, $a=1$.

3) For $3$ coming on two occasions, the two occasions are already covered, because $a=1$ and the first dice has two $2$s. Hence there is no $2$.

4) $4$ comes on three occasions. Now, two of these are covered, because $a=1$ and there are two threes in the first dice. The third case is covered by $b=3$ and the one on the first dice.

5) Five comes on four occasions. Two occasions are $b=3$ and the two $2$s. Another occasion is $a=1$ and the $4$ on the first dice. For one more occasion, let $c=4$, then we are done with the fourth situation, namely $c=4$ and the $1$ on the first dice.

6) Six comes on five occasions. Two of these are covered by $c=4$ and the two $2$s on the first dice. Two more are covered by $b=3$ and the two 3s on the first dice. To cover one more case, we must either have $d=5$ or $d=2$ (for 1 or 4 on the first dice respectively). We take $d=5$, because we ruled out $2$ being on the second dice, in the third step above.

7) Seven comes on six occasions. Two are covered by $d=5$ and the two 2s on the first dice. Two are covered by $c=4$ and the two 3s on the first dice. One case is covered by $b=3$ and the 4 on the first dice. Now, to cover the last case, set $e=6$, that gives you the last case, namely $e=6$ and $1$ on the first dice.

Now, the dice reads $1,3,4,5,6,8$.

Let us separate out the cases to make sure:

$2 \rightarrow 1+1$

$3 \rightarrow 2+1,2+1$

$4 \rightarrow 3+1,3+1,1+3$

$5 \rightarrow 4+1,2+3,2+3,1+4$

$6 \rightarrow 3+3,3+3,2+4,2+4,1+5$

$7 \rightarrow 4+3,3+4,3+4,2+5,2+5,1+6$

$8 \rightarrow 4+4,3+5,3+5,2+6,2+6$

$9 \rightarrow 4+5,3+6,3+6,1+8$

$10 \rightarrow 4+6,2+8,2+8$

$11 \rightarrow 3+8,3+8$

$12 \rightarrow 4+8$

That's the same distribution! Hence your second dice should have $1,3,4,5,6,8$.

Furthermore, by the above procedure, I have shown your answer is unique, so that answers a question in the comments as well.
Hint: We can assume $a \le b \le c \le d \le e \le f$ Start at the ends. You need $\frac 1{36}$ chance to get $2$ and $12$, so $a=1, f=8$ Now think about the chances you need to get numbers like $3,4,10,11$ and see what that tells you about the numbers.