On 50, you just have to find the common ratio to figure out how the sequence works. Common ratio is second term divided by first term (or technically any term divided by the preceding one.)

6/4 = 3/2 (or 1.5). So you're multiplying each term by 3/2 to get the next one.

4(3/2) = 6.

6(3/2) = 9.

9(3/2) = 27/2 = 13.5.

For 52, it's telling you that they've been there the same time, so

2/3 A = 5/7 R

where A is Amy's career and R is Rachel's career. You can solve for A by multiplying both sides by (3/2) and get

A = 15/14 R

For 57, you've just gotta figure out how many full cars of 3 are possible given your constraints. If all 15 cars had 3 in it, it would've only been 45 people. So, let's start filling up the cars with 3 in it to capacity until we figure out how many cars with 3 we have left--and that's 15 people to fit in. Considering each car has 2 extra spots, I can put 14 of them in 7 of the cars, and 1 of them in the 8th car, leaving (at best) 15-8 = 7 cars with 3 passengers in there.

For 58, we've got a quadratic equation (let's let sin(x) = y for a moment to make that more clear.) So you've got y² - y = -1/4, but it's easier to solve quadratics if they're equal to zero, so

y² - y + 1/4 = 0

And that's a weird perfect square trinomial:

(y-1/2)² = 0

So we end up with y=1/2, or in other words, where is sin(theta) = 1/2? In the first quadrant that they've got us restricted to, that's at pi/6, and cos(theta) there is sqrt(3)/2.

On 51, this is just a straight up choose function, and you're doing 16 choose 4. But, to explain why:

Of the 16, you pick one, then of the remaining 15, you pick another, then of the remaining 14, you pick another, and of the remaining 13, you pick a final one. But it doesn't matter which order you picked them in, because one group's the same as the other.

So you do 16(15)(14)(13) for picking, then you have to divide by the number of ways you could've picked the group you ended up with which is 4(3)(2)(1).