Is particle in 1D box, finite & infinite well same case?

Is the potential well with infinite depth the same as the 1D box with infinite potential outside?

Yes, are two different names for the same system, namely the free particle$^dagger$ on the Hilbert space $L^2([-frac{a}{2},frac{a}{2}])$.

For a finite well, the potentials

$$V_1(x)=cases{0 & insideV_0 & outside}$$ and $$V_2(x)=cases{-V_0 & inside 0 & outside}$$

correspond to precisely the same physics. In classical mechanics, this corresponds to being able to freely choose the zero point of the potential energy. In quantum mechanics, this corresponds to being able to multiply all of the energy eigenstates by an overall phase factor, c.f. my answer here.

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In reality, the "infinite potential well" is not defined to be a limiting case of either potential. Instead, it is simply a free particle defined on the Hilbert space $L^2([-frac{a}{2},frac{a}{2}])$ rather than $L^2(mathbb R)$. However, one can loosely interpret the infinite well as a limiting case of the finite well in the sense that all of the bound eigenstates of the finite well approach the corresponding eigenstate of the infinite potential well in the limit as $V_0rightarrow infty$.

Of course, it's possible to poke holes in this interpretation; if we consider the finite well to have potential $V_2$, then the lowest energy eigenvalue goes to $-infty$ in the limit, which doesn't make sense. Furthermore, there would always be unbound states with positive energies, no matter how deep the well gets. For this reason, it's a good idea to bear in mind the crystal-clear point that in reality, the "infinite potential well" is just the free particle on $L^2([-frac{a}{2},frac{a}{2}])$.

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$^dagger$This is well beyond the scope of this question, but in the interest of completeness this is not quite the whole story. It turns out that if $I$ is a compact interval, then the Hamiltonian $H:= -frac{hbar^2}{2m}frac{d^2}{dx^2}$ is not self-adjoint on $L^2(I)$ unless one also provides boundary conditions for the states on which it can act. The choice $psi(-frac{a}{2})=psi(frac{a}{2})=0$ is one possibility, and we call the corresponding system the particle-in-a-box. However, we could make many different choices; choosing $psi(-frac{a}{2})=psi(frac{a}{2})$ (i.e. demanding periodicity) defines the quantum particle-on-a-ring.