= Solution
\b[Weak compactness of the adjoint forces weak compactness of the original operator.] Assume $T^*:Y^*\to X^*$ is weakly compact. Apply the already proved $(i)\Longrightarrow(ii)$ to this operator, rather than assuming the adjoint equivalence in advance. It gives
$$
T^{***}(Y^{***})\subseteq J_{X^*}(X^*).
$$
Take $\psi\in Y^{***}$ annihilating $J_YY$, so $\psi(J_Yy)=0$ for every $y\in Y$. Write $T^{***}\psi=J_{X^*}f$ for some $f\in X^*$. For every $x\in X$,
$$
f(x)=(J_{X^*}f)(J_Xx)=(T^{***}\psi)(J_Xx)
=\psi(T^{**}J_Xx)=\psi(J_YTx)=0.
$$
Hence $f=0$ and $T^{***}\psi=0$. It follows that
$$
\psi(T^{**}F)=0\qquad(F\in X^{**})
$$
for every $\psi$ annihilating $J_YY$.
The <vector subspace> $J_YY$ is closed in the <norm topology> in $Y^{**}$ because $Y$ is complete and $J_Y$ is an <isometry>. The <Hahn-Banach theorem> says that any point outside a closed linear subspace can be separated from it by a <bounded linear functional> vanishing on that subspace. Applying this in $Y^{**}$ shows that an element annihilated by all such $\psi\in Y^{***}$ must belong to $J_YY$. Therefore $T^{**}F\in J_YY$ for every $F$, proving $(iv)\Longrightarrow(ii)$. The reverse implications above now establish all four equivalences, including <weak compactness of an operator and its adjoint>:
$$
\boxed{T\text{ weakly compact}\ \Longleftrightarrow\ T^{**}(X^{**})\subseteq J_YY
\ \Longleftrightarrow\ T^*\text{ is }w^*\text{-to-}w\text{ continuous}
\ \Longleftrightarrow\ T^*\text{ weakly compact}.}
$$
Finally, if $X$ is a <reflexive Banach space>, every $F\in X^{**}$ is $J_Xx$ for some $x$, and $T^{**}F=J_YTx\in J_YY$. If $Y$ is a <reflexive Banach space>, $Y^{**}=J_YY$, so the same range condition holds automatically. In either case, $(ii)\Longrightarrow(i)$ proves
$$
\boxed{X\text{ or }Y\text{ reflexive}\quad\Longrightarrow\quad\text{every bounded }T:X\to Y\text{ is weakly compact}.}
$$
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