The probability of any given event is the number of preferred outcomes divided by the number of all possible outcomes.
When two dice are thrown together, the number of all possible outcomes are 36. You can write them out if you wish e.g. 1,1; 1,2; 1,3; 1,4; 1,5; 1,6; 2,1 etc.
Out of these, the number of preferred outcomes i.e. the number of ways in which a double can occur is 6 i.e. double 1, double 2, double3 .... double 6.
So the probability of a double is 6/36 = 1/6
Since all other outcomes are not doubles, the probability of these is 30/36 = 5/6.
So in answer to the first part of the question, the more likely outcome is that the two dice will show different numbers rather than that they will show the same number.
For the second part we need to consider the probability of a given double being thrown i.e. either 1,1 or 6,6 in this case.
Agai the total number of outcomes when throwing two dice is 36, but now, there is only one way in which we can throw a double 1 i.e. both dice showing 1. So the probability of this happening is 1/36.
In the same way, there is only one way in which a double 6 can be thrown i.e. bothe dice showing a 6 and so the probability of this is also 1/36.
So the probability of throwing a double 1 or a double 6 is exactly the same. Neither is more or less likely.
Hope all that makes sense.
