If 120 buses were 80% of the total, then the total number of scheduled buses can be found by dividing by 0.80: $\frac{120}{0.8}=150$. So, 150 buses were scheduled in March.

Because the cost is proportional to the number of bags, the relationship has the form $c=kb$, where $k$ is the cost per bag. Using the table, divide total cost by number of bags: $13.50÷3=4.50$. Checking with the other row, $22.50÷5=4.50$, so the unit rate is consistent. Therefore, the equation is $c=4.5b$.

The sales increase by $30$ each week because $150-120=30$, $180-150=30$, and $210-180=30$. Since week 1 has sales of $120$, the expression should start at $120$ and then add $30$ for each week after week 1. That is $120+30(n-1)$. Checking: for week 1, $120+30(1-1)=120$; for week 2, $120+30(2-1)=150$. So the correct expression is $120+30(n-1)$.

From the table, every time the number of miles increases by 3, the total fare increases by 6 dollars. That means the fare increases by $\frac{6}{3}=2$ dollars per mile. From 11 miles to 14 miles is another increase of 3 miles, so add 6 dollars to the fare at 11 miles: $27+6=33$. Therefore, the total fare for 14 miles is $33$.

The data show a constant rate of change, so a linear model is reasonable. From 2 to 3 hours, the score increases from 68 to 74, which is an increase of 6 points for 1 hour. From 3 to 5 hours, the score increases by 12 points over 2 hours, which is also 6 points per hour. From 5 to 7 hours, the score increases by 12 points over 2 hours, again 6 points per hour. So the slope is 6 points per hour. A student who studied 6 hours studied 1 more hour than a student who studied 5 hours, whose score was 86. Add 6 points: $86+6=92$. Therefore, the predicted score is 92.