What is the Euler Totient of Zero?

If you read a little further down in the MathWorld entry, you find this:

By convention, $\phi(0)=1$, although Mathematica defines EulerPhi[$0$] equal to $0$ for consistency with its FactorInteger[$0$] command.

However, no reference is given for the convention. It's a rare formula that calls out for $\phi(0)$ to be included. The Book of Numbers, by Conway and Guy, offers one possibility. On page 156, they write:

...the $n$th Farey series has length

$$1+\phi(1)+\phi(2)+\phi(3)+\cdots+\phi(n-1)+\phi(n)$$ (the initial $1$ comes from the fact that we count both $0/1$ and $1/1$).

Note, however, that Conway and Guy do not write $\phi(0)$ for the initial $1$.

In general, we don't often need to define $\phi(n)$ for $n<1$.

For $n=1$, we define $\phi(1)=1$ because we want $\phi$ to be multiplicative, and because it lets us deduce that $\sum_{d\mid n} \phi(d) = n$. The whole reason Wolfram defines it with $\leq n$ rather than $<n$ is just for that case of $n=1$. Essentially, if $\rho$ is any multiplicative function, then $\rho(1)$ is a "empty product," and thus needs to be $1$.

There isn't much value in defining $\phi(0)$. Sometimes, we just leave things undefined. In broader contexts, like algebraic number fields, you can define $\phi(I)$ where $I$ is any ideal in a commutative ring, and $\phi(I)$ is the number of units in $R/I$. That yields a definition of $\phi$ that is more natural and you get $\phi(0)=\phi(0\mathbb Z)=2$. But I don't think that value is useful.

Yes, because we must conclude that there are zero positive integers less than or equal to zero. We do not even have to consider the remainder of the criteria or the fact that $1$ is relatively prime to all numbers.