please help me more thanks ~~~~

1.199p+1=a^2
=>(a+1)(a-1)=199p
p is a prime.199 is also a prime.
=>a+1=199 a-1=p or a+1=p a-1=199
=>p=197,201(not a prime number so it is not a solution)
So B is correct.
2.(X+Y)^2=(X+3)(Y-3)
=>x^2+2xy-(y-3)x+y^2-3(y-3)=0
=>x^2+(3+y)x+(y^2-3y+9)=0
△=(3+y)^2-4(y^2-3y+9)
=-3y^2+18y-27
=-3(y-3)^2
Only when y=3,△=0,there exists x=-3 which can satisfy the equation.
So the answer is B.
3.If(x,y) satisfy the inequation,(-x,y)(x,-y)(-x,-y) will all satisfy the inequation.Therefore,we only need to discuss the cases when x>=0 and y>=0.
x>2,y>2 =>(x+y-4)x+y