Question 1:
y=5^4x-3 what is the ordered pair? (its 5 to the power of 4)
question 2:
which ordered pair makes the equation true?
2y=-5x-4
a. (-2,3)
b. (1,-4)
c. (0,2)
d. (-1,1)

Question 1:
I'm actually not sure which term you ment exactly, so I calculated the solution for three possible variations of your term, depending on how one sets the brackets.

To calculate an ordered pay just set x or y to some value, insert it into the equation and you get the solution for the other variable.
In these calculations I always picked x so values would get eliminated or as close to 0 as possible to make it easier.

variation 1 of term:
y=(5^4)*(x-3)
insert x=3
y=(5^4)*(3-3)==(5^4)*0=0
pair: (3,0)

variation 2 of term:
y=(5^4)*x-3
insert x=0:
y=(5^4)*0-3=0-3=-3
pair: (0,-3)

variation 3 of term:
y=5^(4*x-3)
insert x=1:
y=5^(4*1-3)=5^1=5
pair: (1,5)

question 2:
2y=-5x-4
just insert the pairs as (x,y) into the equation:

a. (-2,3)
2*3=-5*-2-4
6=10-4 yes

b. (1,-4)
2*-4=-5*1-4
-8=-9 no

c. (0,2)
2*2=-5*0-4
4=-4 no

d. (-1,1)
2*1=-5*-1-4
2=5-4 no

12+3b -1=-5-b -12-1+5=-b-3b -8=-4b