Answer: a) 5; b) 5. Solution. a) Note that the area of six different squares is no less than $1+4+9+16+25+36=91>88$. Therefore, it is impossible to cut out more than five different squares. A possible example for five squares (even in a rectangle $8 \times 9$) is shown in the figure. b) Suppose, for the sake of contradiction, that it is possible to cut out 6 squares of different sizes. Then these must be squares of sizes $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4, 5 \times 5$, and $6 \times 6$. After cutting out the $6 \times 6$ square from one side, a strip of width no more than 2 will remain on one side, and a strip of width no more than 6 on the other side. Therefore, the squares $4 \times 4$ and $5 \times 5$ must be located in the strip (no more than) $6 \times 8$. But after cutting out the $5 \times 5$ square from this strip, the remaining part will have strips of width no more than 3, and thus it will be impossible to place the $4 \times 4$ square.
9.2. From a rectangular table of $m \times n$ cells, several squares of different sizes need to be cut out along the grid lines. What is the maximum number of squares that can be cut out if: a) $m=8$, $n=11$; b) $m=8, n=12 ?$