= Solution
For an <irreducible polynomial> $f\in K[X]$, take $L=K[X]/(f)$ and let $\alpha$ be the class of $X$. The quotient is a <field>, is generated by $\alpha$ over $K$, and satisfies $f(\alpha)=0$. Its degree is $\boxed{[L:K]=\deg f}$, with <basis> $1,\alpha,\ldots,\alpha^{\deg f-1}$.
A <splitting field> is a <field> generated over $K$ by all roots of the <polynomial> and in which the <polynomial> is a product of linear factors. Prove existence and the factorial bound together by induction on the degree. Constant nonzero <polynomials> and linear <polynomials> require no extension. For a reducible degree-$n$ <polynomial> write $g=hk$ with positive degrees $r,n-r$. Split $h$ by induction, then split $k$ over that <field>. The resulting <field> is generated by all roots of $g$, and its degree over $K$ is at most $r!(n-r)!\le n!$. If $g$ is irreducible, adjoin one root in degree $n$, divide out its linear factor, and split the degree-$(n-1)$ quotient over that extension. The tower formula gives degree at most $n(n-1)!=n!$. This also works in inseparable <characteristic> and proves the required bound without assuming Galoisness.
For the sextic with constant term three, <Eisenstein's criterion> at three gives degree six. Choose a root $\alpha$, so $\alpha^3$ is a square root of $-3$. Then
$$
\zeta=\frac{1+\alpha^3}{2}
$$
satisfies $\zeta^2-\zeta+1=0$, and is a primitive sixth <root of unity>: $\zeta^3=-1$ and $\zeta\ne-1$. The six distinct roots are $\alpha\zeta^j$ for $0\le j<6$, all in $\mathbb Q(\alpha)$. Hence the simple extension is itself the <splitting field> and $\boxed{[L:\mathbb Q]=6}$.
For the constant term two, Eisenstein at two gives $[\mathbb Q(\gamma):\mathbb Q]=6$ for a root $\gamma$. Its <splitting field> is $\mathbb Q(\gamma,\zeta_6)$. The supplied noncontainment of a root of $X^2+3$ means that adjoining $\zeta_6=(1+\sqrt{-3})/2$ has degree two. Thus
$$
\boxed{[M:\mathbb Q]=12.}
$$
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