Galois extension definition. Let $L,K$ be fields with $L/K$ a field extension. We say $L/K$ is a Galois extension if $L/K$ is normal and separable. 
I don't fully understand this definition, is it saying that 
1) $L$ has to be the splitting field for some polynomial in $K[x]$ and that polynomial must not have any repeated roots, or is it saying that 
2) $L$ has to be the splitting field for all polynomials in $K[x]$ and all polynomials must not have repeated roots?
 A: We define a Galois extension $L/K$ to be an extension of fields that is


*

*Normal: if $x\in L$ has minimal polynomial $f(X) \in K[X]$, and $y$ is another root of $f$, then $y\in L$.

*Separable: if $x\in L$ has minimal polynomial $f(X) \in K[X]$, then $f$ has distinct roots in its splitting field.


When $L/K$ is a finite extension, these conditions are equivalent to $L$ being the splitting field of a separable polynomial $f(X) \in K[X]$ - i.e. your condition $1$. This is a fact which is proven in any course in Galois theory. See for example Theorem 3.10 in these lecture notes.
Your condition $2$ is certainly false: for example $\mathbb Q(\sqrt2)/\mathbb Q$ is a Galois extension, but is not the splitting field of $X^5+3X+2$ or of any other (irreducible) polynomial other than $X^2-2$.
A: A definition is a definition ! Anyway, Roughly speaking , it is saying that $$Gal(L/K)=Aut_K(L)$$
and thus that $Gal(L/K)$ is indeed a group and that $|Gal(L/K)|=[L:K]$. Indeed, if it's just separable, then $Gal(L/K)$ would be $Hom_K(L,K^{alg})$ what is not a group. And if it's only normal, then $$|Aut_K(L)|=|Gal(L/K)|<[L:K].$$
So, finally, since we want that $Gal(L/K)$ be a group (i.e. $Gal(L/K)=Aut_K(L)$,) and that $$|Gal(L/K)|=[L:K],$$
we need to defined $L/K$ as separable and normal. It's thanks to those properties that the correspondance between the (complete) lattice of groups and (complete) lattice of fields make sense. 
A: A separable algebraic extension field $L/K$ is one where given any element $\alpha \in L$, the minimal polynomial $m_{\alpha}(x) \in K[x]$ is separable (that is, has distinct roots in some extension of $K$).
For the most part, the fields one commonly encounters, such as $\Bbb Q, \Bbb R, \Bbb C$ and $\Bbb Z_p$, are separable (these fields are perfect fields, and any finite extension is separable).
In fact, it is not immediately apparent that there are any non-separable fields, so this is the most commonly-given example:
Let $K = \Bbb Z_2(t)$ (this is the field of quotients of the integral domain  $\Bbb Z_2[t]$). Consider the polynomial $x^2 - t \in K[x]$, to which we can assign the splitting field $L = K(\sqrt{t})$. $x^2 - t$ splits in $L$ as $(x - \sqrt{t})^2$ (here we are taking advantage of the fact that $\text{char}(\Bbb Z_2(t)) = 2$). I leave it to you to show $x^2 - t$ is indeed irreducible in $K[x]$ (hint: show we cannot have $t = \dfrac{p(t)^2}{q(t)^2}$ for any $p,q \in \Bbb Z_2[t]$).
In short, your first characterization is better, a separable extension need not be the splitting field for all separable polynomials in $K[x]$.
Normality is a different condition, essentially expressing "how fully" an extension splits an irreducible polynomial. For example, the minimal polynomial of $\sqrt[3]{2} \in \Bbb Q(\sqrt[3]{2})$ in $\Bbb Q[x]$ is $x^3 - 2$.
However, $x^3 - 2$ does not split fully in $\Bbb Q(\sqrt[3]{2})$, but only splits this far:
$x^3 - 2 = (x - \sqrt[3]{2})(x^2 + \sqrt[3]{2}x + \sqrt[3]{4})$,
consequently $\Bbb Q(\sqrt[3]{2})$ is not a normal extension of $\Bbb Q$.
It turns out that for finite extensions there is a one-to-one correspondence between the subfields of $L$ containing $K$ and the subgroups of $\text{Aut}(L/K)$ precisely when $L$ is Galois over $K$ (this is called, surprisingly enough, the Galois correspondence), and that the normal subgroups of $\text{Aut}(L/K)$ then correspond to the normal extension fields of $K$ contained in $L$.
A: the extension $L/K$ is galoisien is equivalently to each $x\in L$, 
$x$ is algebraic over $K$ and the minimal polynomial of $x$ has
simple roots and are all in $L$. so by primitive element theorem
we get: the extension $L/K$  finite and  galoisienne is equivalent
to your proposal 1) but not to 2).
but in the infinity case; $L/K$ galoisienne  is not equivalent
either to 1) or 2)
A: The simpler answer is that your definition (1) is correct in the case $[L:K] < \infty$. 
But let me give a better (generalizable) definition. 
Suppose $K \subset L$ is a field extension.  Then $L$ is said to be Galois over $K$ if   $L^G=K$ where $G=Aut_K(L)$ and $L^G=\lbrace x\in L\mid \sigma(x)=x \ \forall \ \sigma\in G \rbrace$. 
This definition implies your definition (1) in the case $[L:K]< \infty$.  How?
Assuming this definition, if $\alpha\in L$ such that $f(X)\in K[X]$ is the minimal polynomial for $\alpha$, we need to show that $f(X)$ has distinct roots in $L$. Observe that each of $\sigma(\alpha)$ is a root of $f(X)$ since $\sigma$ fixes $K$. 
Thus, the set $T_{\alpha}=\lbrace\sigma(\alpha) \mid \sigma \in G\rbrace$ is finite. 
So we may as well write $T_{\alpha}=\lbrace \sigma_i(\alpha) \in G\mid 1\leq i\leq k \rbrace$ for some positive integer $k$ and $\sigma_i(\alpha)\neq \sigma_j(\alpha)$ for $i\neq j$. 
Consider the polynomial $g(X)=\prod_{\sigma(\alpha)\in T_{\alpha}}(X-\sigma(\alpha))\in L[X]$.  Observe that $\tau(g(X))=g(X)$ for each $\tau\in G$, where the $\tau$ acts on the coefficients of $g(X)$ and leaves $X$ unchanged.  But then our definition of Galois extension implies that $g(X)\in K[X]$.  
By minimality of $f(X)$ we deduce that $f(X)$ divides $g(X)$.  Also, since each $\sigma_i(\alpha)$ is a root of $f(X)$, we deduce that $(X-\sigma_i(\alpha))$ divides $f(X)$ in $L[X]$.   And hence $g(X)\mid f(X)$ in $L[X]$.  It is not hard to digest that $g(X) \mid f(X)$ in $K[X]$.   
Thus, $f(X)=g(X)$.   Then $g(X)$ has distinct roots by definition and thus $f(X)$ is separable. 
Thus, we have shown that every element $\alpha\in L$ is separable.  Using primitive element theorem, we get $L=K(\lambda)$ for some $\lambda \in L$. Let $F(X)$ be the minimal polynomial of $\lambda$.  By above calculation, we observe that $F(X)$ splits in $L$ and  in any field $M$ in which $F(X)$ splits must contain $\alpha$ and its conjugates.  Thus, $L$ is the splitting field of $F(X)$. 
On the other hand, if $L$ was splitting field of $f(X)\in K[X]$. Then it is not hard to show that the Galois group $G$ acts transitively on the roots of any minimal polynomial $g(X)$ of an element $\beta$ in $L$. (It is an argument involving extension of field automorphisms). Thus, if all of Galois group fixes and element $\beta\in L$, it is forced that the minimal polynomial be of degree $1$ and hence we get $L^G=K$. 
