How many sets of two factors of 360 are coprime to each other? My attempt:
$360=2^3\cdot3^2\cdot5^1$
Number of sets of two factor coprime sets for $2^3$ and $3^2$ only $=12+6=18$
With that if we add the effect of $ 5^1$, number of sets $=18+2\cdot 18-1=53$.
Is this ok? 
The answer given is $56$.

I'll assume you mean sets with exactly two elements, so the two factors $1$ and $1$ (represented by the singleton $\{1\}$) won't count. 

Case 1: One of the factors is $1$. 
$\gcd(1,n)=1$ for all $n$, so this is just the number of factors of $360$, minus $1$. The number of factors of $360=2^3\cdot3^2\cdot5$ is simply the product of one plus the exponents on the primes (think of the number of choices for the multiplicity of each prime factor), which is $4 \cdot 3 \cdot 2 = 24$. 
This gives the number of sets in this case as $24-1=23$. 
Case 2: Neither of the factors is $1$, but both are odd. 
This means we choose a power of $3$ and a power of $5$. There are $2$ choices for a power of $3$ and $1$ for a power of $5$ (we don't count $3^0=5^0=1$ in this case). There are $2$ such sets.
Case 3: One of the factors is even, and the other is not $1$. 


*

*If the even number has no odd prime factors, then there are $3$ choices for it: $2, 4,$ and $8$. Given this, there are $3 \cdot 2 -1$ choices for the other factor. This gives $15$ for this sub-case.

*If the even number is divisible by three, then there are $6$ choices for it. Given this, there is only $1$ choice for the other factor: $5$. There are $6$ possibilities in this sub-case.

*If the even number is divisible by five, then there are $3$ choices for it. Given this, there are $2$ choices for the other factor: $3$ or $9$. This gives $6$ possibilities for this sub-case.
There are a total of $15+6+6=27$ possibilites in this case.

Our total is $23+2+27=\boxed{52}$. This doesn't agree with your given answer.
