$A=\{1,2,3,4,5\}$. How many functions $f : A \to A$ so that $(f\circ f)(1) = 3$ 
$A=\{1,2,3,4,5\}$
How many functions $f : A \to A$ so that $f$ is onto?
$5!$

is this correct?

How many functions $f : A \to A$ so that $(f\circ f)(1) = 3$?
$f(1)=1, f(1)=2, f(1)=3, f(1)=4, f(1)=5$?

I don't know what to do...

How many onto functions $f : A \to A$ so that $(f\circ f)(1) = 3$?

 A: Your first answer is correct. As for the second question, consider the following process of constructing a suitable function:
$f(1)$ has to be something.  If $f(1)$ is $1$, then $f(f(1))=1$, which is disallowed.  So, set $f(1)=a$, where $a$ is some number in this set which isn't $1$.
Our constraint is that $f(f(1))=3$.  That is, $f(a)$ has to be $3$.  So, set $f(a)=3$.  We are now sure that our function, however we proceed to construct the rest of it, will satisfy the constraint $f(f(1)) = 3$.  Proceed to extend the function to all other numbers in this set.  Note that any function satisfying the constraint may be constructed in this fashion.
Now, we ask the question: how many ways are there to proceed through this process?
In our first step, setting $f(1) = a$, we have four choices of $a$.  Having chosen an $a$, we have no choice but to set $f(a) = 3$.  We have $3$ more distinct elements $t\in A$ whose value $f(t)$ must be chosen from a set of the $5$ possible options in $A$.
Thus, there are $4\times 5 \times 5 \times 5 = 500$ possibilities.

All right, apparently this process is confusing, so let's build an example function.
First, we need to choose some $a \neq 1$.  Let's say $a = 4$.  Now, we have $f(1) = 4$, and $f(4) = 3$.
With that settled, we have to figure out what $f(2),f(3),$ and $f(5)$ are.  Let's say $f(2) = 1$, $f(3) = 3$, and $f(5) = 4$.  We now have a suitable function.  We could have picked any of $5$ values for each of $f(2),f(3)$ and $f(5)$.
A: The first answer is correct.
Hint : 
$f \circ f (1)=3$   means 
$f(1)=2$ and  $f(2)=3$
or $f(1)=3$ and  $f(3)=3$
or $f(1)=4$ and  $f(4)=3$
or $f(1)=5$ and  $f(5)=3$
