2. If in the septenary system a number is written as $\overline{a b c}=49 a+7 b+c$, then in the system of base 11 it equals $121 c+11 b+a$. Therefore, $49 a+7 b+c=121 c+11 b+a \Longleftrightarrow b=12 a-30 c=6(2 a-5 c)$. Since $b \in[0 ; 6]$, either $b=6,2 a-5 c=1$ (then $c=1, a=3$), or $b=0,2 a-5 c=0$ (then $c=2, a=5$). In the first case, the number is $361_{7}=3 \cdot 49+6 \cdot 7+1=190$, in the second $502_{7}=5 \cdot 49+0 \cdot 7+2=247$.
Answer: 190 and 247.

Answer to option 7-2: 248. (This number is $305_{9}$.)