Splitting fields always exist, given a polynomial $f\in k[X]$ of degree $n$ the field $k_f=k[X]/\lr{f}$ contains a root of $f$ (say $\alpha$), then one can write $f=(x-\alpha)f_1\in k_f[X]$ and now consider the field $k_{f_1}$ (with degree of $f_1$ atmost $n-1$) and so on. Thus there are field extensions $$\cdots[k_{f_1}:k_f][k_f:k]=\cdots (n-1)n$$ Hence, induction implies that $[L:k]\leq n!$, where $L$ is the splitting field of a polynomial $f\in k[X]$ of degree $n$.

Let $\omega$ denote the third root of unity, that is $\omega^2+\omega+1=0$

The splitting field of $x^2+x+1$ is $\Q(\omega)$ and it factorizes as $(x-\omega)(x-\omega^2)$.

The splitting field of $x^2-x+1$ is $\Q(\omega)$ and it factorizes as $(x+\omega)(x+\omega^2)$.

The splitting field of $x^6-1$ is $\Q(\omega)$ and it factorizes as. $$ \begin{align*} x^6-1&=(x^3-1)(x^3+1)\\ &=(x-1)(x^2+x+1)(x+1)(x^2-x+1)\\ &=(x-1)(x-\omega)(x-\omega^2)(x+1)(x+\omega)(x+\omega^2) \end{align*} $$

If $a$ is not a square in $k$, the splitting field of $X^2-a\in k[X]$ is $k(\sqrt{a})$ and the factorization is $(X-\sqrt{a})(X+\sqrt{a})$

If $f(X)=X^p-1\in \Q[X]$, it is known that $\zeta=\exp(2\pi\ci/p)$ is a root and $\Q[\zeta]$ is the splitting field and the factorization is $$X^p-1 =(X-1)(X-\zeta)(X-\zeta^2)\cdots(X-\zeta^{p-1})$$

The above seems natural, if $L=k(\alpha_1,\ldots,\alpha_n)$ and $L'=k'({\alpha'}_1,\ldots,{\alpha'}_n)$, simply extend $\theta$ to each of the roots.
Proof Consider $f\in k[X]$ as irreducible, then correspondingly $g:=\theta(f)\in k'[X]$ is also irreducible (else factorization of $g$ will imply corresponding factorization of $f$). Now, following standard recipe construct fields $k_f$ and ${k'}_g$ with roots designated as $\alpha,\beta$ respectively, that is $k_f=k(\alpha)$ and ${k'}_g=k'(\beta)$, then extend $\theta$ to send $\alpha$ to $\beta$.
  Write $f=(X-\alpha)f_1\in k_f[X]$ and $g=(X-\beta)g_1\in {k'}_{g}[X]$ and repeat the procedure by considering $k_{f_1}$ and ${k'}_{g_1}$ and so on, inducting on the degree of the polynomial.