Question #1 is really bothering me. I can't solve it and this is for middle schoolers? I'm so out of practice with math...
The problems were very likely never meant to be serious problems, and were originally written as a joke. But to answer it seriously, let's take a look at the question:
Johnny has an AK-47 with an 80-round clip. If he misses 6 out of 10 shots and shoots 13 times at each drive-by shooting, how many drive-by shootings can he attempt before he has to reload?
Note that he shoots 13 bullets at each drive-by, and note that we want to find out how many drive-bys he can do before his 80 bullet clip empties and he needs to reload. The fact that he misses 6/10 shots is completely irrelevant. Therefore, it's a simple division problem. Divide the size of his clip by the number of shots per drive-by (80/13) and we get 6 drive-bys with a remainder of 2 bullets left in the clip. There's the answer.
The key is discovering that his missing 6/10 shots is irrelevant (which is a thing that some word problems do). However, let's assume that the question were worded that he needs to HIT 13 targets at each drive-by instead of SHOOT 13 bullets. This question is more complex because you need to take into account the hit percentage.
In that case, because he misses 6/10 (60%) of the time, we know he hits 4/10 (40%) of the time. We can use this to get the number of shots required to hit any target by cross multiplying: 13/x = 4/10, or x = 13 * 10 / 4, which is 32.5. So for each drive-by, to compensate for his inaccuracy, he needs to shoot 32.5 shots to hit 13 targets.
Then we can divide the clip size by the number of shots (80/32.5) which equals 2 drive-bys with 15 shots left in the clip. And there's your answer to the harder problem.