$I^{\alpha }f(x)=\frac{1}{\Gamma(\alpha ) }\displaystyle\int_{a}^{x}f(t)(x-t)^{\alpha -1}dt$
In this definition of the Riemann–Liouville integral, wikipedia says that "$a$ is an arbitrary but fixed base point". And later it says that "The dependence on the base-point $a$ is often suppressed, and represents a freedom in constant of integration".
In some examples I found that $a$ becomes $0$, $-\infty$, $1$, etc...
I really wanna know what is the exact definition of this "$a$".
Do you know the meaning of the word "arbitrary"? The word "arbitrary" refers to the fact that which specific value of $a$ you choose literally does not matter (as long as $a$ is a real number: things change if considering $a\to\infty$ or $a\to-\infty$). In practice, which value you choose depends on the application and on context. So contrary to what you are suggesting, you are not restricted to choosing $a$ to be a zero of the function you are integrating, though it necessarily will be a zero of the antiderivative you are choosing to work with.
However, I should point out that if you are integrating the function $$f(t)=\frac1{t},$$ then you cannot choose $a\to-\infty,$ because then the integral diverges. You should choose $a=1$ in that case.
You asked about the definitoion of $a,$ but in asking this, you are missing the point. $a$ is arbitrary. It is not defined as anything. You can choose it to be whatever you want it to be, so long as it is in the domain of the function being integrated. So we typically just choose whatever we find most convenient. But we are not obligated to work with a specific value of $a.$ Choice is a matter of convenience, not a matter of mathematical requirement or definition.
You asked about the case where $F$ has no zeroes. But that is nonsensical. By definition, $F$ will always have a $0.$ If $f$ is integrable, then $$F(t)=\int_a^tf(x)\,\mathrm{d}x$$ will always have a zero at $a.$ And again, $a$ is arbitrary. In the case of the function $f(x)=-\cot(x)\csc(x),$ you can have $F(t)=\int_a^t-\cot(x)\csc(x)\,\mathrm{d}x=\csc(t)-\csc(a),$ and the value of $a$ is arbitrary: you can choose whatever value of $a$ you want, as long it lies in the same interval $t$ lies in (otherwise, the integral does not exist). Typically, though, the choice that tends to be most convenient in most applications is $a=\frac{\pi}2.$ So typically, you are working with $F(t)=\csc(t)-1,$ not $F(t)=\csc(t).$ There does not exist a situation in which $F(t)=\csc(t),$ it is an impossible situation, since $\csc(a)$ can never be $0.$ So $F$ will always have a zero at $a.$