2. Let $A=\overline{a b c b a}$ be a five-digit symmetric number, $a \neq 0$. If $1 \leq a \leq 8$, then the last digit of the number $A+11$ will be $a+1$, and therefore the first digit in the representation of $A+11$ should also be $a+1$. This is possible only with a carry-over from the digit, i.e., when $b=c=9$. Then $A+11=(a+1) 999(a+1)$ is a symmetric number for any $a=1,2, \ldots, 8$. The case $a=9$ is impossible, since $A+11$ ends in zero, and thus, due to symmetry, it should start with zero. But a number cannot start with zero.

The total number of solutions is equal to the number of possible choices for the number $a$, which is eight.
Answer: eight numbers of the form $\overline{a 999 a}$, where $a=1,2, \ldots, 8$.