= Solution
\b[The bidual range condition implies both weak compactness and the stronger continuity of $T^*$.] First assume $T^{**}(X^{**})\subseteq J_YY$. The <Banach-Alaoglu theorem> makes $B_{X^{**}}$ weak-star compact, and $T^{**}$ is weak-star continuous. Hence its image is compact for the <weak-star topology> in $Y^{**}$. Because it lies in $J_YY$, it corresponds to a weakly compact subset $C\subseteq Y$ under $J_Y^{-1}$.
The identity $T^{**}J_X=J_YT$ gives $T(B_X)\subseteq C$. A <weakly compact set> in a <Hausdorff space> is closed for the <weak topology>, so $C$ is also norm-closed and contains the norm closure of $T(B_X)$. That closure is a <convex set> closed in the <norm topology>, hence weakly closed by the <Mazur theorem>. It is therefore a weakly closed subset of the weakly compact $C$, and is a <weakly compact set>. This proves $(ii)\Longrightarrow(i)$.
For $(ii)\Longrightarrow(iii)$, fix $F\in X^{**}$. By the range assumption there is $y_F\in Y$ with $T^{**}F=J_Yy_F$. For every $y^*\in Y^*$,
$$
F(T^*y^*)=(T^{**}F)(y^*)=y^*(y_F).
$$
The right side is a defining weak-star continuous evaluation on $Y^*$. Since the <weak topology> on $X^*$ is determined by all $F\in X^{**}$, every scalar coordinate of $T^*$ for that topology is weak-star continuous. Therefore
$$
\boxed{T^*:(Y^*,w^*)\longrightarrow(X^*,w)\text{ is continuous}.}
$$
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