Just to flesh out Bill's answer and comments thereon, we have the following facts. Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator.
- If $T$ has dense range then $T^*$ is injective.
Since this is a standard homework problem I'll just give a hint. Suppose $f \in Y^*$ with $T^*f=0$. This means that $f(Tx)=0$ for every $x$, i.e. $f$ vanishes on the range of $T$...
- Suppose further that $X$ is reflexive. If $T$ is injective then $T^*$ has dense range.
Hint: By Hahn-Banach, to show that $T^*$ has dense range, it suffices to show that if $u \in X^{**}$ vanishes on the range of $T^*$, then $u=0$. And by reflexivity, $u \in X^{**}$ is represented by some $x \in X$...
Note that the proofs I have in mind don't need to discuss the weak-* topology (at least not explicitly).
We cannot drop reflexivity in 2. Consider the inclusion of $\ell^1$ into $\ell^2$. It is injective, but its adjoint is the inclusion of $\ell^2$ into $\ell^\infty$, whose range is not dense.
(Fun fact: you can't prove 2 without Hahn-Banach or some other consequence of the axiom of choice. If you work in ZF + dependent choice (DC), it's consistent that $\ell^1$ is reflexive, but we still have the injective map from $\ell^1$ to $\ell^2$ whose adjoint doesn't have dense range. So under those axioms it's consistent that 2 is false.)