1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive

No offence but neither is the approach to this problem correct nor is the solution.

Lets start with the first statement:

I. x^2+y^2>z^2
Squaring both sides we get:
x^4+y^4+2x^2y^2 > z^4
Since 2x^2y^2 is a positive quantity, on it's own this expression is insufficient to determine whether x^4+y^4>z^4

Alternatively,
Let's try and plug-in certain values in this expression:
a. x=1; y=0; z=0
x^2+y^2>z^2
Clearly, x^4+y^4>z^4 holds true here since [((0+1)^2)^2]>[(0)^2]^2
TRUE

b. x=1; y=3; z=2
z^2=4
x^2+y^2=(1+9)=10
Clearly x^2+y^2>z^2

x^4=1; y^4=81; z^4=16
x^4+y^4 > z^4
TRUE

c. x=4; y=5; z=6
z^2=36
x^2+y^2=(16+25)=41
Clearly x^2+y^2>z^2

x^4=256; y^4=625; z^4=1296
x^4+y^4 < z^4
FALSE

Hence, Statement I is INSUFFICIENT

Let's look at the second statement:
II. x+y>z

Squaring both sides
(x+y)^2>z^2
x^2+y^2+2xy > z^2

Squaring again we get:
x^4+4x^3y+6x^2y^2+4xy^3+y^4 > z^4

This expression is Insufficient on it's own to determine whether x^4+y^4 > z^4 since don't know whether x,y,z are positive or negative integers.

Alternatively,
Lets plug-in certain values for x,y and z:
a. x=1; y=1; z=0
(1^4 + 1^4) > 0^4
Hence, x^4+y^4 > z^4
TRUE

b. x=2; y=3; z=4
(2^4+3^4)= (16+81)=97
4^4=256
x^4+y^4 < z^4
FALSE

c. x=1; y=1; z=-10
(x^4+y^4)=2
z^4= 10000
x^4+y^4 < z^4
FALSE

Hence, Statement II is INSUFFICIENT

Combining both the statements doesn't help either as there is no common ground between the two statements which could lead to a solution to the question.

Hence the Answer is E