Two events

Event $3:$ Sum of scores is even. Event $4:$ sum of scores is odd.

There are three odd and three even numbers between $1, 2, 3, 4, 5, 6$ . Since $\text{even}+\text{even}=\text{even}=\text{odd}+\text{odd}$ , from the $6*6=36$ ways, $3*3+3*3=18$ ways are even, which is equal to $36-18=18$ . So Event $3$ and $4$ have the same probality.

Event $1:$ Sum of scores is $2:$ . vent $2:$ sum of scores is $12:$ .

The $2$ can only produced by $1+1$ , and $12$ only by $6+6$ , so they have the same probality.

Event $5:$ Sum of scores is prime. Event $6:$ sum of scores is not prime.

The possible prime numbers are $2, 3, 5, 7, 11$ .

$2=1+1 (1), 3=1+2=2+1 (2), 5=1+4=4+1=2+3=3+2 (4), 7=1+6=6+1=2+5=5+2=3+4=4+3 (6), 11=5+6=6+5 (2),$

so the probality is: $\dfrac{1+2+4+6+2}{36}=\dfrac{15}{36}$ , which is not equal to $\dfrac{36-15}{36}$ . So this is the correct answer.

Event $7:$ Sum of scores is even. Event $8:$ sum of scores is odd.

Let's make number pairs: $1-6, 2-5, 3-4$ . The x number's pair is $x'=7-x$ . So if $a+b<7$ $(0<a, b<7)$ , then $a'+b'=14-a-b>7$ . And if $a+b>7$ , then $a'+b'=14-a-b<7$ , so the probality is the same.