We obtain all ordered pairs $(x,y)$ such that $xy=N$ and $\gcd(x,y)=1$ by choosing a subset of the set of prime divisors of $N$, and giving all these primes to $x$, and giving the rest of the primes to $y$. A set of $n$ elements has $2^n$ subsets, so there are $2^n$ ways to produce a suitable $x$, and hence $2^n$ ordered pairs $(x,y)$.
We now have a count of the ordered pairs. But we want to count the number of factorizations as a product of two relatively prime numbers, and for example $(20)(3)$ is considered to be the same factorization as $(3)(20)$. For $n\ge 1$, to count the unordered pairs, divide the number of ordered pairs by $2$. We get $2^{n-1}$. For a more familiar example, there are $(52)(51)$ ordered pairs of $2$ cards, but there are only $(52)(51)/2$ "hands" of two cards.
The only case where we do not divide by $2$ is the case $N=1$, that is, $n=0$, where the only factorization is $(1)(1)$. So the exact result is that the number of factorizations of the desired type is $2^{n-1}$ if $n\ge 1$ and $1$ if $n=0$.
Comment: Consider the product
$$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots \qquad(\ast)$$
Look at a typical full term $1+p+p^2+\cdots +p^k$. We produce the suitable divisors $x$ of $N$ by deciding whether we will use the $1$ or the $p^k$. No other choice is possible, because if we choose $p^i$ where $0<i<k$, then $N/x$ (in our notation, $y$) will be divisible by $p$, so $x$ and $y$ will not be relatively prime.
Thus at every term in the product $(\ast)$, we have $2$ choices, "None" or "All." There are $n$ terms in the product, so there are $2^n$ ways to produce $x$. Now argue as before that if $n \ge 1$, every suitable factorization is produced twice by this process.
Best Answer
Let's assume you are talking about natural numbers greater than one and positive factors.
Lets say that the prime decomposition is $p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_n^{a_n}$ where of course $n\in\Bbb N$, $n\ge 1$, and for all $a_k$, $a_k\in\Bbb N$, $a_k\ge 1$.
All the divisors of this number can be expressed with the same primes and with exponents between $0$ and $a_k$ inclusive. This means the total number of factors is $(a_1+1)(a_2+1)\ldots (a_n+1)$.
If that number is odd, the original number was a square. A factorization that you want uses two of the factors, except for the expression of the number as a square which uses only one. Thus the number of desired factorizations is
$$\frac{(a_1+1)(a_2+1)\ldots (a_n+1)+1}2$$
If that number is even, the original number is not a square, and we get the number
$$\frac{(a_1+1)(a_2+1)\ldots (a_n+1)}2$$
Your last question is not clear.