For me, in this type of question you have to imagine all the different possibilities and try them to check if you can deduce something from the hypothesis 1 and 2. The thing is not to forget any possibility and to find counter-examples by plugging in, just focus on this kind of DS if you are weak on it.
Is m+z>0 ?
1)m-3z>0 so m>3z
Let's pick number randomely to test the hypothesis.
If m=-1 and z=-2 we respect 1) and we have m+z<0
If m=5 and z=1 we respect 1) and we have m+z>0
Therefore, 1 is insufficient.
2) 4z-m>0 so 4z>m
We pick number as well.
If z=1 and m=0 we respect 2) and m+z>0
If z=-1 and m=-8 we respect 2) and m+z <0
Again, we can conclude either m+z>0 or m+z<0 so 2) is insufficient.
1) and 2)
m>3z
4z>m
So 3z<m<4z and 3z<4z
3z<4z tells us that z is a positive number and consequently m as well.
So m+z>0
And the answer is C
I think the point is 3z<4z enables us to state that z>0
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