Comparing the zero point vibrational energies of DCl and HCl
I don't know which equations you are using, but this is the one I was taught:
$$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$
Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity constant:
$$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$
where $D_\mathrm{e}$ is the depth of the lowest point of the potential well (not the depth of the $v = 0$ vibrational level!).
The zero-point energy is the energy when $v=0$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $(D_\mathrm{e})$ of the $\ce{H-Cl}$ and $\ce{D-Cl}$ bonds are same.
The only difference between $\ce{H-Cl}$ and $\ce{D-Cl}$ would be in the value of the vibrational wavenumber $(\tilde{\nu}),$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.