3. Let $c_{n}=11 \ldots 1$ be a number with $n$ ones in its decimal representation. Then $c_{n+1}=10 \cdot c_{n}+1$. Therefore,

$$
c_{n+1}^{2}=100 \cdot c_{n}^{2}+22 \ldots 2 \cdot 10+1
$$

For example,

$c_{2}^{2}=11^{2}=(10 \cdot 1+1)^{2}=100+2 \cdot 10+1=121$,

$c_{3}^{2}=111^{2}=100 \cdot 11^{2}+220+1=12100+220+1=12321$,

$c_{4}^{2}=1111^{2}=100 \cdot 111^{2}+2220+1=1232100+2220+1=1234321$, and so on.

Notice that for all the listed numbers $c_{2}^{2}, c_{3}^{2}, c_{4}^{2}$, the digit around which these numbers are symmetric (2 in the case of $c_{2}^{2}$, 3 in the case of $c_{3}^{2}$, 4 in the case of $c_{4}^{2}$) matches the number of ones in the number being squared. The number $c=123456787654321$ given in the problem is also symmetric around the digit 8, i.e., it can be assumed that it is the square of the number $c_{8}=11111111$. This can be verified by column multiplication or using the recursive relation (*).
Answer: 1) $a^{2}=121 ;$ 2) $b^{2}=12321 ;$ 3) $\sqrt{c}=11111111$.