I think the answer is yes. You may look at any book on spectral theory.

Here is a comment for the dimension 2 case. Let $N=\mathbb{H}\backslash\Gamma$ be a hyperbolic surface, where $\mathbb{H}$ is the upper half-plane and $\Gamma$ is a finite-index subgroup of $SL(2,\mathbb{Z})$. Here $\Gamma$ acts on $\mathbb{H}$ by Mobius transformation. It is well-known from the spectral theory (see Lang, $SL(2,\mathbb{R})$, for example) the spectrum bellow 1/4 is discrete. Now take $\Gamma'$ be a finite index subgroup of $\Gamma$. Then the projection map $\pi:\mathbb{H}\backslash\Gamma'\rightarrow\mathbb{H}\backslash\Gamma$ is a finite cover. Moreover, by the above, the spectrum of $\mathbb{H}\backslash\Gamma'$ is still discrete bellow 1/4.

I think the compact case is even simpler since the spectrums are all discrete.

I think the answer is yes. You may look at any book on spectral theory.

Here is a comment for the dimension 2 case. Let $N=\Gamma\backslash\mathbb{H}$ be a hyperbolic surface, where $\mathbb{H}$ is the upper half-plane and $\Gamma$ is a finite-index subgroup of $SL(2,\mathbb{Z})$. Here $\Gamma$ acts on $\mathbb{H}$ by Mobius transformation. It is well-known from the spectral theory (see Lang, $SL(2,\mathbb{R})$, for example) the spectrum bellow 1/4 is discrete. Now take $\Gamma'$ be a finite index subgroup of $\Gamma$. Then the projection map $\pi:\Gamma'\backslash\mathbb{H}\rightarrow\Gamma\backslash\mathbb{H}$ is a finite cover. Moreover, by the above, the spectrum of $\Gamma'\backslash\mathbb{H}$ is still discrete bellow 1/4.

I think the compact case is even simpler since the spectrums are all discrete.