Переводим числа из указанных систем счисления в десятичную.
\( A01_{16} = A × 16^2 + 0 × 16^1 + 1 × 16^0 \)
\( A_{16} = 10_{10} \)
\( A01_{16} = 10 × 256 + 0 × 16 + 1 × 1 \)
\( A01_{16} = 2560 + 0 + 1 \)
\( A01_{16} = 2561_{10} \)
\( 162_{7} = 1 × 7^2 + 6 × 7^1 + 2 × 7^0 \)
\( 162_{7} = 1 × 49 + 6 × 7 + 2 × 1 \)
\( 162_{7} = 49 + 42 + 2 \)
\( 162_{7} = 93_{10} \)
\( 213_{4} = 2 × 4^2 + 1 × 4^1 + 3 × 4^0 \)
\( 213_{4} = 2 × 16 + 1 × 4 + 3 × 1 \)
\( 213_{4} = 32 + 4 + 3 \)
\( 213_{4} = 39_{10} \)
Ответ: A0116 = 256110, 1627 = 9310, 2134 = 3910.