Can the null set be an element of a set? The cartesian product, $\{\varnothing\} \times \{\text{piano, flute}\}$, gives:
$\{(\varnothing,\text{piano}),(\varnothing,\text{flute})\}$.
But if, in $(a,b)$, either a or b $= \varnothing$, then the pair $= \varnothing$.
So product is really $\{\varnothing, \varnothing\}$ or $\mathbb{\{\varnothing\}}$. Am I correct that this product is different from $\varnothing$, meaning that a set containing $\varnothing$ is different from $\varnothing$ alone?
 A: Yes, the null set can be an element of a set. 
You are correct that a set containing $\varnothing$ is different from $\varnothing$ alone; for example: $\{\varnothing\} \ne \varnothing$. Or as a concrete example, the latter is equivalent to an empty baggie, while the former is a baggie containing an empty baggie.
But it is incorrect to say that "if, in $(a,b)$, either $a$ or $b$ $= \varnothing$, then the pair $= \varnothing$"; that's simply not true.
A: No, indeed: $\big\{\{\}\big\}\times \big\{a,\{\}\big\} = \big\{(\{\}, a), (\{\}, \{\})\big\}$ and neither pair is an empty set, nor is the Cartesian product empty.
Where as: $\big\{\big\}\times\big\{a,\{\}\big\} = \big\{\big\}$ 
The empty set by itself is not the same thing as a construct containing the empty set.
$$\{\varnothing\}\neq \varnothing\\ (\varnothing, x)\neq \varnothing$$
A: Yes. For each set $X$, we have $\emptyset \in P(X)$, the power set of $X$.
A: I suspect that when you wrote

if, in $(a,b)$, either $a$ or $b=\varnothing$, then the pair $=\varnothing$

you may have actually had in mind the following statement, which (unlike the quote above) is actually true:

If, in $a \times b$, either $a$ or $b=\varnothing$, then the Cartesian product $=\varnothing$

If I am right -- if that is actually what you meant -- than I suspect you are having trouble distinguishing in a consistent way between the members of a set and the set itself.  $a \times b$ is not an ordered pair, but a collection of ordered pairs, each of the form $(x,y)$ where $x\in a$ and $y\in b$.  If $a$ has no elements, that is if $a=\varnothing$, then you can't form any ordered pairs whose first member is an element of $a$.  But you can still form an ordered pair whose first element is $a$ itself.
A: Yes, it is an element of the set of subsets of every set.
https://en.wikipedia.org/wiki/Power_set#Example
