$V_k$ being a model of ZFC whenever $k$ is strongly inaccessible 
ZFC implies that the $V_k$ is a model of ZFC whenever $k$ is strongly
  inaccessible..

So if $k$ is weakly inaccessible, it can't be a model of ZFC? Why is it like this?

And ZF implies that the Godel universe $L_k$ is a model of ZFC
  whenever $k$ is weakly inaccessible.

Again, the same question.
 A: First thing we should note that $L$ satisfies GCH, so if $\kappa$ is weakly inaccessible it is strongly inaccessible in $L$. This will be important later.
Now, to see that $V_\kappa$ is a model of ZFC if $\kappa$ is inaccessible we need to note several things:


*

*If $x\in  V_\kappa$ then $|x|<\kappa$, this is because $\kappa$ is inaccessible so $\lambda<\kappa\implies 2^\lambda<\kappa$, and so we can prove that for all $\alpha<\kappa$, $|V_\alpha|<\kappa$ (this is by transfinite induction).

*Extensionality holds immediately, so does regularity and infinity axioms. The power set holds in $V_\alpha$ for any limit ordinal, so it holds here as well. In fact all but union and replacement axioms hold.

*Union holds because if $A\in V_\kappa$ we have that $|A|<\kappa$, and if $A\in V_\kappa$ then all the members of $A$ are in $V_\kappa$, so their cardinality is $<\kappa$. Since $\kappa$ is regular we have that whenever we take a union of less than $\kappa$ sets of size less than $\kappa$ the result has cardinality less than $\kappa$. It remains to verify that the union is indeed a set in $V_\kappa$, but we can do this by seeing that the rank of the union is the supremum ($+1$ maybe) of the ranks of the members of the members of $A$. All those ordinals are below $\kappa$, so unions are below $\kappa$ and we are fine.

*Replacement holds for the same reason. You need to show that whenever $f$ is a definable function in $V_\kappa$ (perhaps with parameters from there) and $A\in V_\kappa$ then $f''A\in V_\kappa$. 

*Choice is obvious by all the above. Every set in $V_\kappa$ is in bijection with an ordinal below $\kappa$.
Now going back to $L$, show that if $\kappa$ is inaccessible then $L_\kappa=V_\kappa$. Since weakly inaccessible in $L$ is the same as inaccessible we are done.
