There are lots of problems with this question

Given: \((2z)^{-2} = \frac{1}{36}\)

Rewrite as: \(\frac{1}{(2z)^2} = \frac{1}{36}\)

Simplify: \(\frac{1}{4z^2} = \frac{1}{36}\)

From this we can conclude that: \(4z^2 = 36\)

Divide both sides by 4 to get: \(z^2 = 9\)

So, z = 3 OR z = -3


Given: \((x^y) \times (x^2)=x^6\)
One possible solution here is: x = 5 and y = 4, since (5^4)(5^2) = 5^6
Another possible solution is: x = 0 and y = 99, since (0^99)(0^2) = 0^6
Another possible solution is: x = 1 and y = -7, since (1^77)(1^2) = 1^6
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Since y can have infinitely many values, the correct answer here is D

Cheers,
Brent