1. We are given that $\alpha$ is a root of $1+x^2+x^4=0.$ So, $1+\alpha^2+\alpha^4=0.$

2. Rewrite this as $\alpha^4+\alpha^2+1=0.$ Now factor: $\alpha^4+\alpha^2+1=\frac{\alpha^6-1}{\alpha^2-1}$ for $\alpha^2\ne 1$, or more directly, $\alpha^6-1=(\alpha^2-1)(\alpha^4+\alpha^2+1).$ Since $\alpha^4+\alpha^2+1=0$, we get $\alpha^6=1.$ Also, $\alpha^2\ne 1$ because if $\alpha^2=1$, then $1+\alpha^2+\alpha^4=1+1+1=3\ne 0$.

So $\alpha$ is a 6th root of unity, but not $\pm 1$.

3. We need to find $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}.$ Since $\alpha^6=1$, powers of $\alpha$ repeat modulo $6$.

Compute the exponents modulo $6$:

• $1011 \equiv 3 \pmod{6}$, because $1011=6\cdot 168+3$
• $2022 \equiv 0 \pmod{6}$, because $2022=6\cdot 337$
• $3033 \equiv 3 \pmod{6}$, because $3033=6\cdot 505+3$

Hence, $\alpha^{1011}=\alpha^3, \qquad \alpha^{2022}=1, \qquad \alpha^{3033}=\alpha^3.$

4. Substitute these into the expression: $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}=\alpha^3+1-\alpha^3=1.$

5. Therefore, the value is $\boxed{1}.$

6. Option check:

• A: $1$ ✅
• B: $\alpha$ ❌
• C: $1+\alpha$ ❌
• D: $1+2\alpha$ ❌

So the correct option is A.