Prove that $x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$ converges Assume that $x_1$ is a real number and for $n \gt 1$ :
$x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$  
a) Assume that $x_1=0.6 $
Prove that $\{x_n\}$ converges and find its limit.  
b) Find a set of real numbers like $A$ such that if $x_1 \in A$ then $\{x_n\}$ is convergent.   
Note 1 : I know a method to answer the part "a" but its not a precise way. ( Put $L$ instead of every $x_i$ )
Note 2 : I have no idea about part "b".
 A: Here is a graph that complements other two answers:

Starting from the point $(x_1, 0)$, you move upward to the point $(x_1, x_2)$ where $x_2 = f(x_1)$. Here, the function $f$ is defined as
$$f(x) = x + 0.8x(1-x) - 0.0072.$$
Then we update the $x$-coordinate to this new value to obtain $(x_2, x_2)$. This amounts to moving horizontally until you hit the diagonal line $y = x$. Then you again move vertically until you reach $(x_2, x_3)$ with $x_3 = f(x_2)$, and repeat this process.
This picture gives a lot of intuition on the recursion $x_{n+1} = f(x_n)$. For instance, this convinces you that the sequence $(x_n)$ increases and converges to one of the solution of $x = f(x)$. You can improve this intuition to a rigorous observation, which will indeed let you prove the convergence.
For other starting point, notice the difference in behavior:

A: To your Note 1, if you put $L$ instead of the x you will maybe find the potential limits i.e. if it converges for that $a_0$ the limit is among Ls. But in fact it is possible that it does not converge. 
Way to go :
1) Find $f$ such that $x_{n+1}=f(x_n)$
2) Graph f and y=x
3) Write done $a_0$, draw its image on y and draw the value of y on the x axis using the y=x line. Restart with the value (i.e. $a_2$) and see if it seems to converge to on possible L.
You can easily saw the set of such starting point for questions 2.
Use basic analysis to prove your point.
A: Let 
\begin{align}
f(x) &= x+0.8x(1-x)-0.072 \\
 &= x+\frac{4}{5}x(1-x) - \frac{9}{125} \\
 &= \frac{9}{10}+\frac{9}{25}\left(x-\frac{9}{10}\right)-\frac{4}{5} \left(x-\frac{9}{10}\right)^2 
\end{align}
I obtained that last expression by expanding $f$ in a series about $9/10$. You can expand out the last two expressions to see that they are equal. 
Subtracting $9/10$ off from both sides, factoring out an $(x-9/10)$ from the right and taking the absolute values, we get
\begin{align}
  \left|f(x)-\frac{9}{10}\right| &= \left|x-\frac{9}{10}\right|
\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right|.
\end{align}
Thus, the distance between $f(x)$ and $9/10$ will be less than the distance between $x$ and $9/10$ precisely when
$$\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right| < 1.$$
This last inequality is equivalent to 
$$\frac{1}{10}<x<\frac{13}{5}.$$
Thus, the sequence will converge to $9/10$ whenever $x$ is in that interval.
