Yes, because $\: I\Delta_0 + \text{WPHP}\left(\Delta_0\right) \:$ proves the unboundedness of primes (see this answer),
since the assetion that a $\Delta_0$-defined relation is an injection from $\:[0\hspace{.01 in},\hspace{-0.02 in}2\hspace{-0.05 in}\cdot\hspace{-0.04 in}x]\:$ to $\:[0,\hspace{-0.01 in}x]$
can be made itself $\Delta_0$ by modifying the relation to also require that its output is in $\:[0,\hspace{-0.01 in}x]\;$.

Yes, because $\: I\Delta_0 + \text{WPHP}\left(\Delta_0\right) \:$ proves the unboundedness of primes (see this answer),
since the assetion that a $\Delta_0$-defined relation is an injection from $\:[0\hspace{.01 in},\hspace{-0.02 in}2\hspace{-0.05 in}\cdot\hspace{-0.04 in}x]\:$ to $\:[0,\hspace{-0.01 in}x]$
can be made itself $\Delta_0$ by modifying the relation to also require that its output is in $\:[0,\hspace{-0.01 in}x]\;$.

Yes, because $\: I\Delta_0 + \text{WPHP}\left(\Delta_0\right) \:$ proves the unboundedness of primes (see this answer),
since the assetion that a $\Delta_0$-defined relation is an injection to $[0,x]$ can be
made itself $\Delta_0$ by modifying the relation to also require that its output is in $[0,x]$.

Yes, because $\: I\Delta_0 + \text{WPHP}\left(\Delta_0\right) \:$ proves the unboundedness of primes (see this answer),
since the assetion that a $\Delta_0$-defined relation is an injection from $\:[0\hspace{.01 in},\hspace{-0.02 in}2\hspace{-0.05 in}\cdot\hspace{-0.04 in}x]\:$ to $\:[0,\hspace{-0.01 in}x]$
can be made itself $\Delta_0$ by modifying the relation to also require that its output is in $\:[0,\hspace{-0.01 in}x]\;$.