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(B) Let the two sets be $A$ and $B$ with $|A| = m$ and $|B| = n$.The number of subsets of set $A$ is $2^m$ and the number of subsets of set $B$ is $2^

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$6, 3$

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(B) Let the two sets be $A$ and $B$ with $|A| = m$ and $|B| = n$.The number of subsets of set $A$ is $2^m$ and the number of subsets of set $B$ is $2^n$.According to the problem,$2^m - 2^n = 56$.We can factor this as $2^n(2^{m-n} - 1) = 56$.Since $56 = 8 	imes 7 = 2^3 	imes 7$,we compare the powers of $2$ and the odd factors.Thus,$2^n = 2^3$,which implies $n = 3$.Also,$2^{m-n} - 1 = 7$,which implies $2^{m-n} = 8 = 2^3$.Therefore,$m - n = 3$.Substituting $n = 3$,we get $m - 3 = 3$,so $m = 6$.Thus,the values are $m = 6$ and $n = 3$.

Two finite sets have $m$ and $n$ elements. The total number of subsets of the first set is $56$ more than the total number of subsets of the second set. The values of $m$ and $n$ are

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