Sure.

Note first that we have a short exact sequence

$$ 0 \to O_X \xrightarrow{t} O_X \to O_{X_t} \to 0 $$ (if $D$ is the divisor corresponding to $t = 0$ on $X$, you can also view this as $$0 \to O_X(-D) \to O_X \to O_D \to 0.$$

Anyways, now apply the functor $mathcal{H}om_{O_X}(\bullet, \omega_X)$. You get $$ 0 \to \mathcal{H}om_{O_X}(O_{X,t}, \omega_{X}) \to \mathcal{H}om_{O_X}(O_X, \omega_X) \xrightarrow{t} \mathcal{H}om_{O_X}(O_X, \omega_X) \to \mathcal{E}xt^1(O_{X_t}, \omega_X) \to \mathcal{E}xt^1_{O_X}(O_X, \omega_X) $$ Ok, the first term is zero (homing torsion into non-torsion), the last term is also zero since you are homing from a free module. Finally, $\mathcal{E}xt^1(O_{X_t}, \omega_X) = \omega_{X_t}$ by basic properties of dualizing/canonical modules, and the result follows.

Sure.

Note first that we have a short exact sequence

$$ 0 \to O_X \xrightarrow{t} O_X \to O_{X_t} \to 0 $$ If $D$ is the divisor corresponding to $t = 0$ on $X$, you can also view this as $$0 \to O_X(-D) \to O_X \to O_D \to 0.$$

Anyways, now apply the functor $\mathcal{H}om_{O_X}(\bullet, \omega_X)$. You get $$ 0 \to \mathcal{H}om_{O_X}(O_{X,t}, \omega_{X}) \to \mathcal{H}om_{O_X}(O_X, \omega_X) \xrightarrow{t} \mathcal{H}om_{O_X}(O_X, \omega_X) \to \mathcal{E}xt^1(O_{X_t}, \omega_X) \to \mathcal{E}xt^1_{O_X}(O_X, \omega_X) $$ Ok, the first term is zero (homing torsion into non-torsion), the last term is also zero since you are homing from a free module. Finally, $\mathcal{E}xt^1(O_{X_t}, \omega_X) = \omega_{X_t}$ by basic properties of dualizing/canonical modules, and the result follows.