Let $s$ be the number of student tickets and $a$ be the number of adult tickets. The total number of tickets gives

The total cost gives

Substitute $s=25-a$ into the cost equation:

Simplify:

So the school bought $10$ adult tickets.

Let the original width be $w$ feet, so the original length is $w+8$. Since the perimeter is $80$,

So the original dimensions are $16$ by $24$.

The original area is

square feet. Since the walkway area is $264$ square feet, the remaining grass area is

square feet.

Because the walkway is built inside all four sides, the grass rectangle has dimensions

So,

Expanding gives

Divide by $4$:

Factoring gives

So $x=3$ or $x=17$. Since $x=17$ would make the inside dimensions negative, it is impossible. Therefore, the value of $x$ is $3$.

If $x$ is the number of adult tickets, then the number of student tickets is $x+20$. The total amount collected is

Now simplify:

So the equivalent expression is $x(a+s)+20s$.

Let $x$ be the number of $5$-dollar tickets and $y$ be the number of $8$-dollar tickets. The actual revenue is $5x+8y$.

If all tickets had been $5$ dollars, the revenue would be $5(x+y)$, which is $144$ less than the actual revenue:

This simplifies to

so

If all tickets had been $8$ dollars, the revenue would be $8(x+y)$, which is $96$ more than the actual revenue:

This simplifies to

so

Therefore, the total number of tickets sold is

Let $n$ be the number of notebooks and $p$ be the number of pens. The total number of items gives $n+p=41$. The total amount of money gives $3n+2p=94$. Substitute $p=41-n$ into the second equation:

So, $12$ notebooks were sold.