The intersection of 2 CFL
The set $A$ cointains all the strings on the alphabet $\{a,b,c\}$ starting with $a$, ending with $c$ and with the same number of $b$s and $c$s. Similarly, $B$ is made by strings on the same alphabet, starting with $a$, ending with $c$ and with the same number of $a$s and $b$s.
So the intersection $A\cap B$ simply cointains all the strings on the alphabet $\{a,b,c\}$ starting with $a$, ending with $c$, with the same number of $b$s and $c$s and the same number of $a$s and $b$s. By the transitivity of equality (I mean, if $x=y$ and $y=z$ then $x=z$, where $x$, $y$ and $z$ are the numbers of $b$s, $c$s and $a$s in a string), you can express these last two condition by saying that the number of $a$s, $b$s and $c$s is the same, i.e., that the strings in $A\cap B$ are of the form $a^n b^n c^n$ for some natural number $n$ (of course here you can use the letter $m$, or any other letter you prefer, instead of $n$).