(i) Take "(x,y)\\in A\\times B". Then "x\\in A" and "y\\in B", hence by "A\\subset C" and "B\\subset D" we have "x\\in C" and "y\\in D." This implies that "(x,y)\\in C\\times D", which gives "A\\times B\\subset C\\times D".
(ii) By definition, "A\\times B=\\{(a,y),(a,z),(b,y),(b,z),(c,y),(c,z),(d,y),(d,z)\\}" and "B\\times A=\\{(y,a),(z,a),(y,b),(z,b),(y,c),(z,c),(y,d),(z,d)\\}"
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