For the first question, you just need to plug the values and solve for k:
[tex]11 = \dfrac{3000}{500}+k \iff 11=6+k \iff k=11-6=5[/tex]
Now we know that [tex]k=5[/tex], so the cost of 200 cards will be
[tex]y = \dfrac{3000}{200}+5 = 15+5=20[/tex]
Finally, if we want the cost to be 7 we're asking
[tex]7=\dfrac{3000}{x}+5 \iff \dfrac{3000}{x}=2 \iff x = \dfrac{3000}{2}=1500[/tex]
