We can't do tree diagrams on here.  

There are C(7,2) = (7*6)/(2*1) = 21 ways the two top cards can occur.

There are only 3 even numbered cards [2,4,6}, so there are only
C(3,2) = (3*2)/(2*1) = 3 ways both top cards can be even.  They
are {2,4}, {2,6}, and (4,6}. 

So the probability that both will be even is 3/21 or 1/7.

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To have an odd sum, one card must be even and the other odd.

We can pick the odd card 4 ways {1,3,5,7}
We can pick the even card 3 ways (2,4,6}

That's 4*3 = 12 ways to pick them so that the sum will be odd.

So the probability that the sum will be odd is 12/21 = 4/7.

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You'll have to ask your teacher how he wants you to draw
a tree diagram to figure this out, but the answers are 1/7
and 4/7.

Edwin