To answer your question, you need to go back to the definition of fractional exponents. What $a^{1/n}$ really means is the solution to the equation $x^n = a$. From you post I infer that you are not very familiar with complex numbers, so I am going to assume that we are only looking for real solutions. For even $n$, $x^n = a$ has two solutions if $a \geq 0$ and no solutions if $a < 0$. For odd $n$, $x^n = a$ always has exactly one solution. Hence, if $n$ is odd there is no problem to define $a^{1/n}$. However, if $n$ is even, there is a choice to be made and the standard choice is to take the positive solution. Now, knowing $a^{1/n}$ we can easily define $a^{m/n}$ by just multiplying $a^{1/n}$ $m$ times. So, using these definitions, $(-1)^{7/3}$ is actually not a problem. Indeed $(-1)^{1/3} = -1$ so that $(-1)^{7/3} = -1$.
So, what is the issue here? The issue with your calculator is probably that it does not think that way, but rather has an exponentiation function programmed for all real numbers. For real numbers that can not be written as $\frac{m}{n}$, so-called irrational numbers, you need to be more creative to define exponents. The standard way of doing it is $a^x = e^{x\ln(a)}$ because it yields the right result for rational $x = \frac{m}{n}$ and positive $a$ (and also has nice continuity properties). And here you see the issue now. $\ln(a)$ is not defined for negative $a$ and thus if you try to compute $a^x$ this way, you get an error on your calculator.
Note that if you are thinking of complex numbers, things get even more difficult because then the equation $x^n = a$ actually has $n$ solutions and it is a priori not clear at all which to pick. Again you would go and make some conventions, pick positive real solutions for positive real numbers and so on. The message here is, for non-integer exponents, everything depends on conventions and different conventions lead to different results (and frequent posts of $-1 = 1$ "proofs" posted here on math.stackexchange).
