Let me answer your second question first.
The weak$^{\ast}$-topology is Hausdorff (let me treat the real case, the complex case is similar): If $\phi \neq \psi$ are two linear functionals then there is $x \in X$ such that $\phi(x) \lt r \lt \psi(x)$. The sets $U = \{f \in X^{\ast} \,:\,f(x) \lt r\}$ and $V = \{f \in X^{\ast}\,:\,f(x) \gt r\}$ are weak$^{\ast}$-open (since evaluation at $x$ is weak$^{\ast}$-continuous) and disjoint neighborhoods of $\phi$ and $\psi$, respectively.
That the weak topology is Hausdorff is shown similarly, using Hahn-Banach.
Next, if $X$ is separable, then the unit ball in the dual space is metrizable with respect to the weak$^{\ast}$-topology: pick a countable dense set $\{x_{n}\}_{n \in \mathbb{N}}$ of the unit ball of $X$ and verify that
\[
d(\phi,\psi) = \sum_{n=1}^{\infty} 2^{-n} \frac{|\phi(x_n) - \psi(x_n)|}{1+|\phi(x_n) - \psi(x_n)|}
\]
defines a metric compatible with the weak$^{\ast}$-topology. Hence the unit ball is sequentially compact in the weak$^{\ast}$-topology (this can be shown directly using Arzelà-Ascoli, by the way).
Using a standard Baire category argument, one can show that weak$^{\ast}$-compact sets are norm-bounded: Indeed, if $K$ is weak$^{\ast}$-compact, it is a Baire space. Write $B^{\ast}$ for the closed unit ball in $X^{\ast}$. Clearly $K = \bigcup_{n = 1}^{\infty} (K \cap n \cdot B^{\ast})$, so at least one of the closed subsets $K \cap n \cdot B^{\ast}$ of $K$ must have non-empty interior. By compactness finitely many translates of $n\cdot B^{\ast}$ must cover $K$, thus $K$ is bounded in norm and hence $K$ is a closed subset of a large enough ball.
Conclusion: If $X$ is separable then every weak$^{\ast}$-compact subset of $X^{\ast}$ is sequentially compact.
I don't know if the converse is true.
If $X$ is not separable, then weak$^{\ast}$-compactness does not imply weak$^{\ast}$-sequential compactness, the standard example is mentioned in Florian's post.
Since you might be interested in the weak topology as well, there's a rather difficult result due to Eberlein:
Recall that a space is countably compact if every countable open cover has a finite subcover. A sequentially compact space is countably compact.
Theorem (Eberlein) If a subset of a Banach space is weakly countably compact then it is weakly compact and weakly sequentially compact.
and finally:
Theorem (Eberlein-Šmulian) A bounded subset of a Banach space is weakly sequentially compact if and only if it is weakly compact. In particular, if the unit ball is weakly sequentially compact then $X$ is reflexive.
Daniel Fischer said it all. I would like to make a precision which would have been a comment if I had enough reputation.
The key point is that the closed unit ball of $X^*$ is compact for the weak*-topology. That's called Banach-Alaoglu theorem.
Now to get your equivalence, you also need the following easier facts:
the weak*-topology is Hausdorff.
a compact in a Hausdorff space is closed (and Hausdorff is needed).
a closed subset of a compact space is always compact.
a norm bounded set in a normed vector space is contained in a positive-scalar-multiple of the closed unit ball.
to show that weak*-compact implies norm bounded, you need the Uniform Boundedness Principle and the fact that the image of a compact space by a continuous complex-valued map (point evaluations in this case, which are continuous by definition of the weak*-topology) is compact in $\mathbb{C}$, whence bounded.
Best Answer
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(\Bbb N)$ is precisely the set $E=\{ \alpha e_i:i\in\Bbb N, |\alpha|=1 \}$ where $e_i=(0,0,\dots,0,1,0,0,\dots)$, consisting of $0$ for all but the $i^{\text{th}}$ coordinate.
Proof: If $e_i = (1-\lambda)x+\lambda y$ for some $x,y\in B_{l^1}$, then $(1-\lambda)x_i+\lambda y_i=1$ and $|x_i|,|y_i|\le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $\alpha e_i$.
On the other hand, any point $x\in B_{l^1}\backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $x\in \text{co}(E)$, $x$ can have only finitely many nonzero terms, so $\text{co}(E) \subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $\overline{\text{co}} ^*(E)=B_{l^1}$ so $\text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $\text{co}(E)=\text{co}(\text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $\overline{\text{co}}(E)=B_{l^1}$:
We clearly have $$ B_{l^1} = \overline{\text{co}}(E) \subset \overline{\text{co}} ^*(E) $$ since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion $$ \overline{\text{co}} ^*(E) \subset B_{l^1} $$ comes from the fact that $\text{co}(E) \subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that $$\overline{\text{co}} ^*(E)=B_{l^1}.$$