I want to show that:
1) $X^4+1$ is irreducible
The roots are the elements of $$\left\{\frac{\pm 1+i}{\sqrt 2},\frac{\pm 1-i}{\sqrt 2}\right\}$$
therefore it's not a product of a polynomial of degree 1 and a polynomial of degree 3. Is there an easier way to prove that it's not of the form $(X^2+aX+1)(X^2+bX+1)$ than to solve the system $(X^2+aX+1)(X^2+bX+1)=X^4+1$ ?
2) $X^4-16X^2+4$ is irreducible
It's easy to prove that there is no integer roots, therefore it's not a product of a polynomial of degree 1 and a polynomial of degree 3. I suppose that $$X^4-16X+4=(X^2+aX+b)(X^2+cX+d)$$ $$\iff X^4-16X^2+4= X^4+(a+c)X^3+(b+d+ac)X^2+(bc+da)X+bd$$
$$\iff\begin{cases}a+c=0,\quad(1)\\ b+d+ac=-16, \quad (2)\\ bc+da=0,\quad (3)\\ bd=4,\quad (4)\end{cases}$$
By $(4)$ we have that $(b,d)\in\{(1,4),(-1,-4), (2,2), (-2,-2)\}$. By $(1)$, $a=-c$. By $(3)$, $bc+da=bc-dc=c(b-d)=0$, therefore $c=0$ or $b=d$. If $c=0$, then by $(2)$, $b+d=-16$ and no elements of $\{(1,4),(-1,-4), (2,2), (-2,-2)\}$ satisfy this equation, then $c\neq 0$. Therefore $b=d$, and by $(4)$, $b^2$=4 therefore $b=d=\pm 2$. But by $(2)$, $$2b-c^2=-16\implies c^2=16+4=20\implies c=\pm\sqrt{20}\notin \mathbb Z,$$ therefore, the polynomial is irreducible on $\mathbb Q$.
It looks ok ? is there an easier proof ?
3) $X^4+X^2+\frac{1}{2}$
I have no idea. Actually, the real question is compute the degree of the extension $\mathbb Q(i,\alpha)/\mathbb Q$ where $\alpha$ is a root of $X^4+X^2+\frac{1}{2}$. There is maybe an other way to solve the problem. If it can help, the root of this polynomial are the elements of $$\left\{\frac{i}{\sqrt 2}\sqrt{1-i},-\frac{i}{\sqrt 2}\sqrt{1-i},\frac{i}{\sqrt 2}\sqrt{1+i},-\frac{i}{\sqrt 2}\sqrt{1+i}\right\}$$
$1)$ : $$f(x)=x^4+1$$ $$f(x+1)=x^4+4x^3+6x^2+4x+2$$
and the criterion of Eisenstein shows that $f(x+1)$ is irreducible. Therefore $f(x)$ is also irreducible.