This is a well-known property of the Hilbert-space adjoint operator that it is WOT continuous. Is a similar true for Banach spaces? That is, for a given Banach space $X$ is the operation
$\varphi\colon B(X)\to B(X^*)$, $\varphi(T)=T^*$
continuous with respect to WOT-topologies of $B(X)$ and $B(X^*)$? I guess it will be the case for $X$ reflexive.
No. (Of course it does work for reflexive spaces, as noted by Martin.)
Example.
$X = c_0$, $X^* = l_1$, $X^{**} = l_\infty$.
$T_n : c_0 \to c_0$ is the left-shift by $n$, so that $$ (a_1,a_2,\dots) \mapsto (a_{n+1},a_{n+2},\dots) $$ where we deleted the first $n$ coordinates.
Compute $T_n^* : l_1 \to l_1$. It is the right-shift by $n$, so that $$ (a_1,a_2,a_3,\dots) \mapsto (0,0,\dots,0,a_1,a_2,\dots) $$ where we inserted $n$ extra zeros at the beginning.
Then: $T_n \to O$ in the WOT for $B(c_0)$, where $O$ is the zero operator. If $x=(x_1,x_2,\dots) \in c_0$ and $y = (y_1,y_2,\dots) \in l_1$, then $$ \left|\langle T_nx , y\rangle\right| = \left|\sum_{k=1}^\infty x_{k+n} y_k\right| \le \|y\|_1 \left(\max_{j \ge n+1} |x_j|\right) \to 0 . $$
But $T_n^*$ does not converge to $O$ in WOT for $B(l_1)$. Indeed, take $y \in l_1$ with positive coefficients, and $z = (1,1,\cdots) \in l_\infty$. Then $$ \langle z,T_n^* y\rangle = \sum_{k=n+1}^\infty y_{k-n} z_k = \|y\|_1 $$ for all $n$ and does not go to zero.