What exactly is an $n$-th root in a field? What is meant when one says that a finite field contains an $n$th root of $a$? Is it an exquisite way to say that the multiplicative group of the field has an element $x$ such that $x^n=a$? If $a=-1$, is it equivalent to saying that the group contains an element of order $2n$?
What if the field is infinite?
 A: Yes. An $n$-th root of $a$ is some solution of $x^n = a$. The field could be finite or infinite. 
An $n$-th root of $1$ is some solution of $x^n=1$. For example, it could be $1$. There is no implicit meaning that an $n$-th root of $1$ actually has order $n$. An $n$-th root of $-1$ need not have order $2n$. For example, a $3$rd root of $-1$ is the name for a solution of $x^3 = -1$.  One such solution is $-1$, which in characteristic not $2$ has order $2$, not $2n = 6$.
A: Supplementing the other answers somewhat because the OP's comment to KCd's answer brought up the need for this.
We can say a bit more!
Undoubtedly you have seen the following result when studying cyclic groups.
Fact 1. If $c$ has order $n$, then the order of $c^k$ is given by the formula
$$\operatorname{ord}(c^k)=\frac{n}{\gcd(n,k)}.$$
This leads to the following (I have used it many times when answering a question about finite fields). 
Fact 2. Assume that $p$ is a prime, $x^p=a$, and that $k=\operatorname{ord}(a)$ is divisible by $p$. Then
$$\operatorname{ord}(x)=pk.$$
Proof. Let $m=\operatorname{ord}(x)$.  The first fact tells us that
$$
k=\frac{m}{\gcd(m,p)}.
$$
Because $p$ is a prime that $\gcd$ is either $1$ or $p$, so we have either $k=m$ or $k=m/p$. Or, equivalently $m=k$ or $m=pk$. But, $p\mid k$, so
$$
x^k=(x^p)^{k/p}=a^{k/p}\neq1.
$$
Therefore $m\neq k$, so $m=pk$. QED.
Example. If $K$ is a field of characteristic $\neq2$ (so that $-1\neq1$), and $a^2=-1$, then $a$ has order $4$.
So the first fact always constrains the orders of roots of an element of a known order. We similarly see that a cubic root of an element of order $9$ has order $3\cdot9=27$. But, a cubic root of an element of order $8$ has order either $8$ or $24$. In fact, both possibilities occur as you can easily verify in the field of complex numbers. Cubing permutes the solutions of $z^8=1$, so one cubic root of an element of order eight also has order eight.
A: Let $F$ be a field. Let $a \in F$. Let $n \in \Bbb N$.
We say that $F$ has an nth root of $a$ if there is $x \in F$ such that $x^n = a$ (we do not require it to be in the multiplicative group).
If $a = -1$ and $F$ is not of characteristic $2$, then if $x^n = -1$ then we know that the order of $x$ divides $2n$. It does not mean that $x$ has order $2n$ (e.g. $x=-1$, $n=3$).
It does not matter if the field is finite or infinite.
