 Solution

A monic quadratic polynomial over a field is irreducible exactly when it has no root in that field. Testing the nine choices of its two coefficients in ${\mathbb{F}}_3$ givesFor clarity, a zero constant term gives a root at zero. Among constant term one, only the zero linear coefficient avoids roots at $1,2$. Among constant term two, precisely the linear coefficients one and two avoid roots. This exhausts all choices.

Put $f=x^4+x+2$. Its values at $0,1,2$ are respectively $2,1,2$, so it has no linear factor. If this quartic were reducible, absence of a linear factor would force a factorization into two irreducible quadratics. Division by the three candidates above gives remainders respectively $x,x+1,x+1$, all nonzero. Thus $f$ is an irreducible polynomial.

The quotient by an irreducible polynomial is a field consequently gives a field of $3^4=81$ elements. Its nonzero elements form a multiplicative group of a finite field of order $80$. In this field, $a^4=1+2a$. Successive squaring givesThe latter is not one: otherwise the nonzero polynomial $1+x+2x^3$ of degree below four would vanish at $a$, contradicting its minimal polynomial. By Lagrange's theorem, $(a^{16}{)}^5=a^{80}=1$. Since five is prime, its order is exactly five. Hence one answer isIts other nonidentity powers are the remaining primitive fifth roots. It is unnecessary to prove that $a$ itself generates the entire multiplicative group.