If $A=A^{\star}$ is a densely-defined selfadjoint linear operator on a complex Hilbert space $H$, and if there is a complete orthonormal basis of $H$ consisting of eigenvectors of $A$, then it is true that the point spectrum $\sigma_{p}(A)$ of $A$ is dense in $\sigma(A)$. The converse is not true.
To show this, let $H$ be a Complex Hilbert space and suppose $A:\mathcal{D}(A)\subseteq H\rightarrow H$ is a densely-defined selfadjoint linear operator. Suppose $H$ has an orthonormal basis $\{ e_{\alpha} \}_{\alpha \in \Lambda}$ of eigenvectors of $A$ with corresponding eigenvalues $\{\lambda_{\alpha}\}_{\alpha\in\Lambda}$, which must be real. If $x\in\mathcal{D}(A)$, then
$$
Ax = \sum_{\alpha\in\Lambda}(Ax,e_{\alpha})e_{\alpha}=\sum_{\alpha\in\Lambda}(x,Ae_{\alpha})e_{\alpha}
= \sum_{\alpha\in\Lambda}\lambda_{\alpha}(x,e_{\alpha})e_{\alpha}.
$$
If $\mu \in \sigma_{p}(A)$, then $Ax=\mu x$ for some $0\ne x \in \mathcal{D}(A)$, which gives
$$
0=(\mu x- Ax)=\sum_{\alpha\in\Lambda}(\mu-\lambda_{\alpha})(x,e_{\alpha})e_{\alpha}.
$$
Because $x \ne 0$, then $(x,e_{\alpha})\ne 0$ for some $\alpha\in\Lambda$, which forces $\mu=\lambda_{\alpha}$. So
$\sigma_{p}(A)\subseteq\{ \lambda_{\alpha} : \alpha \in \Lambda\}$, while the opposite inclusion is obvious. Hence, $\sigma_{p}(A)=\{\lambda_{\alpha} : \alpha\in\Lambda\}$.
The spectrum $\sigma(A)$ is closed; so $\sigma_{p}(A)^{c}\subseteq \sigma(A)$ (here 'c' denotes topological closure.) To see that $\sigma_{p}(A)^{c} =\sigma(A)$, we assume $\lambda \notin\sigma_{p}(A)^{c}$ and show that $\lambda\notin\sigma(A)$. To show this, note that if $\lambda\notin\sigma_{p}(A)^{c}$, then there exists $\delta > 0$ such that
$$
|\lambda_{\alpha}-\lambda| \ge \delta > 0,\;\;\; \alpha\in\Lambda.
$$
Therefore, if $x \in \mathcal{D}(A)$,
$$
\|(A-\lambda I)x\|^{2}=\sum_{\alpha}|\lambda_{\alpha}-\lambda|^{2}|(x,e_{\alpha})|^{2} \ge \delta^{2}\|x\|^{2}.
$$
Because $A=A^{\star}$, the above is enough to prove that $\lambda\notin\sigma(A)$, which proves that $\sigma_{p}(A)^{c}=\sigma(A)$.
To see that the converse is false, let $H=L^{2}[0,1]\times L^{2}[0,1]$ and let $\{ e_{n}\}_{n=1}^{\infty}$ be a complete orthonormal subset of $L^{2}[0,1]$. Define
$$
A(f,g) = (tf(t),\sum_{n=1}^{\infty}q_{n}(g,e_{n})e_{n}),
$$
where $\{ q_{n}\}$ is an enumeration of the rational numbers in $[0,1]$. Then
$$
A(0,e_{n})=q_{n}(0,e_{n}),
$$
which implies that $\{ q_{n}\}\subseteq \sigma_{p}(A)$. And $A(f,g)=\lambda (f,g)$ implies $f=0$ and $g=e_{n}$ for some $n$. So, even though $\sigma_{p}(A)^{c}=[0,1]=\sigma(A)$, $H$ cannot have a basis of eigenvectors for $A$.
One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by
$$(S_nx)_m = \left\{ \begin{matrix}
\frac{x_m}{m} & m \leq n \\
0 & m > n
\end{matrix}
\right.
$$
Note that the $S_n$ are finite-rank and that
$$((T-S_n)x)_m = \left\{ \begin{matrix}
0 & m \leq n \\
\frac{x_m}{m} & m > n
\end{matrix}
\right.
$$
so $\lVert T-S_n\rVert_2^2 = \frac{1}{n+1} \to 0$ as $n \to \infty$. Hence, $T$ is compact. By the Fredholm Alternative, the non-zero spectrum of $T$ consists purely of eigenvalues (i.e. the point spectrum). Since the spectrum must be closed, it must also contain $0$; as $T$ is injective and bounded, $0$ cannot belong to the point or residual spectra so it belongs to the continuous spectrum.
Best Answer
The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has $$ \begin{align} (-\Delta -\frac{1}{\|x\|})f & = -\Delta f -\frac{1}{r}f \\ & = -\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{df}{dr}-\frac{1}{r}f. \end{align} $$ There is a purely radial eigenfunction of this operator which corresponds to the ground state of the Hydrogen atom. This eigenfunction has the form $f(r)=e^{-r/2}$ in this case where the physical constants are missing: $$ \begin{align} (-\Delta-\frac{1}{r})e^{-r/2} & = -\frac{1}{r^{2}}\frac{d}{dr}(r^{2}(-\frac{1}{2})e^{-r/2})-\frac{1}{r}e^{-r/2} \\ & = -\frac{1}{4}e^{-r/2}+\frac{1}{r}e^{-r/2}-\frac{1}{r}e^{-r/2}=-\frac{1}{4}e^{-r/2}. \end{align} $$ Therefore $-1/4 \in \sigma_{P}(-\Delta-\frac{1}{|x|})$. There are lots of other negative eigenvalues and eigenfunctions for this operator.
I already answered the first part of your question in the comment section. The spectrum of the Fourier transform is $\{ 1, i, -1, -i \}$.
Final Note: I see that you changed the question again. This was a complete and valid answer to the original question, and the answer is not a trivial one. Before that, the last paragraph above was a complete answer to the question. HAMMER, if you don't like the answer, it's best to ask a separate question, rather than invalidate a perfectly good answer that someone has put thought and effort into. You obviously learned something from my answers, and it's not polite to invalidate that in such a way that it makes my answer look completely irrelevant to the topic.