Answer
$a.\quad (3k+1)^{2}=3(3k^{2}+2k)+1$
...(one more than a multiple of 3)
$b.\quad $ No. $\quad (3k+2)^{2}=3(3k^{2}+4k+1)+1$
...(one more than a multiple of 3, not two more)
Work Step by Step
a. If $m=3k+1$...(one more than a multiple of 3)
$m^{2}=(3k+1)^{2}=$
$=9k^{2}+6k+1=$
$=3(3k^{2}+2k)+1$...(one more than a multiple of 3)
...which proves the statement
b. If $m=3k+2$...(two more than a multiple of 3)
$m^{2}=(3k+2)^{2}=$
$=9k^{2}+12k+4=$
$=9k^{2}+12k+3+1=$
$=3(3k^{2}+4k+1)+1$...(one more than a multiple of 3, not two more)
...which disproves the statement.
