here is a hint: the order of the kernel of such a homomorphism has to be a common divisor of 36 and 24. this kernel is a subgroup of Z36. the possible common divisors of 36 and 24 are the possible divisors of 12, namely: 1,2,3,4,6 and 12.

a property of cyclic groups is: if d is a divisor of the order of a cyclic group G, G has EXACTLY one subgroup of order d.

{0}, the unique subgroup of Z36 of order 1.

{0, 18}, the unique subgroup of Z36, of order 2.

{0, 12, 24}, the unique subgroup of order 3.

{0, 9, 18, 27}, the unique subgroup of order 4.

{0, 6, 12, 18, 24, 30} the unique subgroup of order 6.

{0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33}, the unique subgroup of order 12.

where can you go from here?