Give an isomorphism of the field with $F_2(\gamma)$ where gamma solves $x^6+x+1$. $F_2(\alpha, \beta)$ where $\alpha$ is a root of $x^2+x+1$ and $\beta$ is  a root of $x^3+x+1$.
Give an isomorphism of this field with  $F_2(\gamma)$ where gamma solves $x^6+x+1$.

$$
\begin{aligned}
F_2(\alpha)&=\{ a_1 +a_2 \alpha :a_1 ,a_2 \in F_2 \}
\\ F_2(\beta)&=\{ a_1 +a_2 \beta +a_3 \beta^2 :a_1 ,a_2 \in F_2 \}
\end{aligned} $$
Not 100% sure about this
$$F_2(\alpha,\beta)=\{ (a_1 +a_2 \alpha) +(a_3 +a_4 \alpha )\beta +(a_5 +a_6 \alpha)\beta^2: a_1,a_2 ,a_3,a_4,a_5,a_6 \in F_2\}  $$
My understanding is that  $ F(\alpha,\beta )$ has something to do with double extension 
$$(F_2[x]/(x^3+x+1))/(x^2+x+1) $$
That extension can be made in sage where a lot of meat grinding can be done 
Fp=GF(2)
p.<x>=PolynomialRing(Fp)
f=x^2+x+1
f2=x^3+x+1
f3=x^6+x+1
Fp2=Fp.extension(f2,'a')
Fp3=Fp2.extension(f,'a')
Fp4=Fp.extension(f3,'a')
Fp4.cardinality()
Fp3.cardinality()

for a in Fp4:
    if ((a^3+a+1==0)  ): 
        print(a)


output 


    a^3 + a^2 + a
    a^4 + a^3 + 1
    a^4 + a^2 + a + 1


for a in Fp4:
    if ((a^2+a+1==0)  ): 
        print(a)


a^5 + a^4 + a^3 + a + 1
a^5 + a^4 + a^3 + a

The original plan was to find an element in $F_2(\alpha,\beta)$ where its the root of $x^6+x+1$ and make that a mapping that should be an isomorphism. 
Not sure where to go from here

In any field of characteristic $p$ you have the Frobenius automorphism (and its powers). So here, you have all $\phi_k(x)=x^{(2^k)}$ which works. In view of your plan, if $\gamma$ is s solution of $x^6+x+1=0$ then the image of $\gamma$ is $\phi_k(\gamma)=\gamma^{(2^k)}$ which, for small $k$ ($0\leq k\leq 5$), are the distinct solutions of $x^6+x+1=0$. In fact, $\phi_0=Id$ and all $\phi_k$ are different for $0\leq k\leq 5$.   
