y²+8y+k

Suppose it factors as

(y+a)(y+b) where a and b are positive integers.

Then FOIL-ing that out,

y²+by+ay+ab

Factor y out of the middle two terms:

y²+(b+a)y+ab

Compare that to 

y²+8y+k

So the coefficient of y which is b+a=8, and the last term ab=k

b   a   b+a   ab=k       factorization
1   7    8     7=k    y²+8y+7 = (y+1)(y+7)
2   6    8    12=k   y²+8y+12 = (y+2)(y+6)
3   5    8    15=k   y²+8y+15 = (y+3)(y+5)
4   4    8    16=k   y²+8y+16 = (y+4)(y+4) = (y+4)²

Answer: all positive integral valuse for k are 7,12,15, and 16

Now you do the other one the same way.

Edwin