That idea leads you to the solution.
If the characteristic is $5$ you get $\gcd$ $=1$.
Now assume that characteristic is not $5$. Then you can divide by $5$ (cleaner to multiply by $5$) and compute the gcd.
We get, $\gcd(x^5-x^2+1,5x^3-2)=\gcd(5x^5-5x^2+5,5x^3-2)=\gcd(3x^2-5,5x^3-2)$
If characteristic is $3$ we get the $\gcd=1$. Assume characteristic is not $3$ either.
We have that $3017=7\cdot 431$. For these primes we get $\gcd\neq1$.
Actually, we get $\gcd=25x-6$. This tells us what the possible multiple root would be.
Let us do the characteristic $7$ case. What this $\gcd$ is telling us is that the solution of $25x-6=0$, i.e. $x=5$, is a root of our polynomial and its derivative.
Divide $x^5-x^2+1$ by $x-5$. We get $x^5-x^2+1=(x-5)^2(x^3+3x^2+5x+2)$.
Using the old definition of separable:
In this case we ($char =7$) we have $x-5$ is an irreducible factor with only one root. We will need to do with the factor $x^3+3x^2+5x+2$, the same study we did for the original polynomial.
Computing the $\gcd$ with its derivative (we proceed as before, multiplying by leading coefficients to ease the division, and taking remainder mod $7$)
$\gcd(x^3+3x^2+5x+2,3x^2+6x+5)=\gcd(3x^3+2x^2x+6,3x^2+6x+5)=\gcd(3x+1,3x^2+6x+5)=\gcd(x+4,0)=x+4$. So $x+4$ is a multiply factor of $x^3+3x^2+5x+2$. To not have to compute any further, notice that because this polynomial is of degree $3$ it should split completely. So, I guess it would be called separable according to the old definition (if I am reading it correctly).
The case of $char=431$ is all yours. But the work is just the same.