For problem 2):
Scoring 7 points means (presumably) the sum of the rolls is 7. Imagine you roll a die twice, in succession. You record the results as an ordered pair giving the result of the first roll, then the result of the second roll.
A score of 7 occurs in (and only in) the following ways: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).
Out of the 36 equally likely outcomes that result from throwing a die twice in succession, exactly six will have a score of 7. So the probability of scoring 7 is ${6\over 36}={1\over6}$.
See Dilip' comment below. With unordered outcomes it's twice as likely to get "a one and a three" as "two threes".
Let's solve the problem using indistinguishable outcomes:
The outcomes from rolling two dice (dies?) simultaneously are $ \matrix{ \{1,1\} & \{1,2\}& \{1,3\}&\{1,4\}& \{1,5\} &\{1,6\} \cr & \{2,2\}&\{2,3\}& \{2,4\} &\{2,5\} &\{2,6\}\cr & &\{3,3\}& \{3,4\} &\{3,5\} &\{3,6\}\cr & & & \{4,4\} &\{4,5\} &\{4,6\}\cr & & & &\{5,5\} &\{5,6\}\cr & & & & &\{6,6\}\cr } $
Now, imagine, please, that the two dice were of different colors, say taupe and ecrue.
In how many ways could you have obtained the outcome $\{1,2\}$? The answer is two ways: taupe was was 1 and ecrue was 2, or taupe was 2 and ecrue was 1.
In how many was could you have obtained the outcome $\{2,2\}$? Well, only one way: both of the dies showed "2".
A moment's reflection should convince you that an outcome that has different entries is twice as likely to occur as an outcome that has the same entries.
Further reflection reveals that any two outcomes of the "different entry" variety have the same probability of occurring and any two outcomes of the "same entry" variety have the same probability of occurring.
From this, we can calculate probabilities of individual outcomes: let the probability of an outcome with same entries be $a$. Then the probability of an outcome with different entries is $2a$.
Then, since the sum of the probabilities of all outcomes is 1 $ 1= 6\cdot a+15\cdot (2a). $
So $a=1/36$, and we have
$ P(\text{outcome with same entries})={1\over36} $ and $ P(\text{outcome with different entries})={2\over36} $
Now we can find
$ P( \text{score of 7} ) =P(1,6)+P(2,5)+P(3,4) =3\cdot {2\over36}={1\over6}. $
That was quite a bit of reflecting to arrive at the answer. As far as "real life" is concerned, you could imagine the two dice (dies?) were distinguishable; and in that case, it is much easier to convince yourself that there are 36 equally likely outcomes.