1. The extension $\Q(\sqrt[n]{2})/\Q$ is not normal for $n\geq 3$.
2. The extension $\Q(\sqrt[3]{2},\omega)/\Q$ (where $\omega$ is third root of unity) is normal, since it is a splitting field of $X^3-2\in\Q[X]$ (see theorem below).

Proof Let $K/k$ a finite normal extension, and let $(x_1,\ldots,x_d)$ a base of $k$ vector space $K$, and let $P_i\in k[X]$ be the minimial (irreducible) polynomial of $x_i$. Since, $K/k$ is normal each $P_i$ splits in $K$, then so does $Q:=P_1\ldots,P_d$. As $K$ is generated on $k$ by the $x_i$, that are the roots of $Q$, the field $K$ is a splitting field of $Q\in k[X]$.
  Let $K$ be the splitting field of the polynomial $Q\in k[X]$, then it is a finite extension of $k$. Let $P\in k[X]$ be an irreducible polynomial that has a root $x_1\in K$, and let $L$ (containing $K$) be the splitting field of $P$, and let $x_2$ be a root of $P$ in $L$. It suffices to show that $x_2\in K$, since this implies that all the roots of $P$ in $L$ are in fact in $K$, thus showing $P$ already splits in $K$.
  For each $i\in\{1,2\}, K(x_i)$ is a splitting field of $Q$ on $k(x_i)$. On the other hand, $k(x_i)=k[X]/P$, that is there is a $k$ isomorphism $\sigma:k(x_1)\ra k(x_2)$. The extensions $k(x_1)\hookrightarrow K(x_1)=K$ and $k(x_1)\simeq k(x_2)\hookrightarrow K(x_2)$, which gives $[K:k(x_1)]=[K(x_2):k(x_2)]$ giving $K=K(x_2)$ (using tower law of field extensions). Hence, $x_2\in K$.

Remark: If $L/k$ is a finite and normal field extension, and say $K$ is an intermediate field between $k$ and $L$. The theorem implies that the extension $L/K$ is again normal. However, it is not the case for the extension $K/k$, as shown by the example $\Q(\sqrt[3]{2},\omega)\supset \Q(\sqrt[3]{2})\supset\Q\quad(L/K/k)$.
  Furthermore, the composition of two normal extensions $K/k$ and $L/K$ is not necessarily normal, as demonstrated by the example $\Q(\sqrt[4]{2})\supset\Q(\sqrt{2})\supset\Q\quad(L/K/k)$.

Proof: If the extension $K/k$ is normal $K$ is the splitting field of a polynomial $Q\in k[X]$ (by the theorem above). For all $k$ homomorphisms $\sigma:K\ra\Omega$, the image of $\sigma$ is the extension of $k$ generated by the roots of $Q$ in $\Omega$, so does not depend on $\sigma$.
  Inversely, suppose that all the $k$ homorphisms of $K$ in $\Omega$ have the same image, that we denote by $K'\subset\Omega$. Let $P\in k[X]$ be an irreducible polynomial with a root $x\in K$ and let $y$ be a root of $P$ in $\Omega$. The fields $k(x)\subset K$ and $k(y)\subset\Omega$ are $k$ isomorphic since $k(x)=k[X]/P=k(y)$. Hence, we can extend $k$ homomorphism $k(x)\ra k(y)$ to $K\ra\Omega$, and the image of $K$ is $K'$ by assumption. We deduce that $y$ is in $K'$ (since the mapping $k(y)\ra K'$ is injective). $$ \require{AMScd} \begin{CD} K @>{\simeq}>> K'@>{\subset}>>\Omega\\ @AAA @AAA \\ k(x) @>{\simeq}>> k(y) \end{CD} $$ The polynomial $P$ splits in $K'$, thus in $K$ since the two $k$ extensions are isomorphic. Hence the extension $K/k$ is normal.

Proof We already know by corollary 1 above that $K/k$ is normal if and only of $\tau(K)=K$ for $\tau$ a $k$ homomorphism of $K$ in $\Omega$. Here we are given $\tau$ as a $k$-automorphism of $L$. Thus, we need to show $k$ homomorphism of $K$ induces $k$ automorphism of $L$ and vice-versa, which will imply the result.
  Let $\Omega$ be an algebraically closed field containing $L$ (exists by Steinitz theorem). By CL1T theorem all $k$-homomorphism of $K$ in $\Omega$ extends to $L$ and since $L/k$ is normal, the image in $\Omega$ of this extension is $L$ (by corollary 1 above); it thus induces a $k$-automorphism of $L$. To summarise, if $L/k$ is normal, all $k$ homomorphisms of $K$ induce $k$ automorphisms of $L$.
  Inversely, all $k$-automorphism $\tau$ of $L$ induce a $k$-homomorphism of $K$ in $\Omega$.

Proof Let $(x_1,\ldots, x_d)$ be a the base of $k$ vector spaces $K$ and let $P_i\in K[X]$ be the minimal polynomial of $x_i$. Let $L\subset\Omega$ be the sub field generated by the roots of $Q:=P_1\cdots P_d$ in $\Omega$. It is the splitting field of the polynomial $Q$, thus the extension $L/k$ is finite and normal.
  Moreover, for all fields $K\subset L'\subset \Omega$ such that the extension $L'/k$ is normal, the irreducible polynomial $P_i$ has a root in $L'$ (that is $x_i$), thus it splits (definition of normal extension), hence so does $Q$. We deduce that $L$ that is generated by the roots of $Q$ is contained in $L'$.