We throw two dices distinc. What is the probability that there will be at least a 6 given that the dice show different results? We throw two dices distinc. 
What is the probability that there will be at least a 6 given that the dice show different results?

My work

Let 

$A=$We have a six, 

$B=$Different result

Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{6}$

This because: $P(B)=\frac{30}{36}$ and $P(A\cap B)=\frac{5}{36}$

Is good this?
Observe that the numbers on the two dice are distinct. They are also indistinguishable, so that any one of the six numbers has equal probability of showing. The probability that one of them is a six is therefore one third (two numbers out of six are showing).
I agree until your calculation for $P(A\cap B)=\frac{5}{36}$. Note that A or B could be the $6$, and that we want them to not both be 6. There's $12$ ways that A or B is $6$ and $1$ way that they are both $6$, so $P(A\cap B)=\frac{12 - 1}{36}=\frac{11}{36}$.

Substitute this back in and the final answer would be $\frac{11}{30}$