Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
Let $A$ stand for the (dense) set of $(a_n)_n$. Since $\sigma(T)$ is closed and $A\subseteq\sigma(T)$, it follows that $[0,1]=\overline{A}\subseteq\sigma(T)$. On the other hand, if $\lambda\notin[0,1]$, then $S$ defined by $Se_j=\frac{1}{\lambda-a_j}e_j$ for any $j\in\mathbb{N}$, is a bounded (diagonal) operator whose inverse is $\lambda I-T$.Thus $\sigma(T)=[0,1]$.
If $\mu$ is an eigenvalue, then we have $Tx=\mu x$ for some non-zero $x=\sum b_je_j$. Then $Tx=\sum a_j b_j e_j=\sum \mu b_j e_j$. Hence, for any $j\in\mathbb{N}$, $a_j b_j=\mu b_j$, which only can happen if $\mu=a_k$ for some $k$. Thus $\sigma_p(T)=A$.
The generalization has almost identical proof. Take $A$ to be a countable, dense subset of $K$, and define a diagonal operator as above.
Best Answer
To answer Q1, we can say exactly when a subset of $\mathbb{C}$ can be realized as the point spectrum of a bounded linear operator on $\ell^2$.
In fact, we can go a little bit further. If $K$ is contained in the open ball $B_R(0)$ for some $R > 0$ then it is possible to choose $T$ such that $\Vert T\Vert=R$.
Before constructing $T$ with the given point spectrum, I'll quickly prove the converse. If $K=\sigma_p(T)$ then $K$ is a bounded Fσ set (this will use the fact that $\ell^2$ is a Banach space whose dual is separable). First, $K$ is contained in the spectrum $\sigma(T)$, which is contained in the closed ball $\bar B_R(0)$ for $R=\Vert T\Vert$ (by standard properties of bounded operators), so is bounded. Next, let $B=\lbrace x\in \ell^2\colon\Vert x\Vert\le 1\rbrace$ be the closed unit ball in $\ell^2$ with the weak topology, which is compact. Then, $B^0\equiv B\setminus\lbrace0\rbrace$ is σ-compact. In fact, choosing any dense sequence $u_1,u_2,\ldots$ in $\ell^2$, we can explicitly write $B^0$ as the union of compact sets $$ B^0=\bigcup_{n=1}^\infty\lbrace x\in B\colon\vert\langle u_n,x\rangle\vert\ge1\rbrace. $$ Set $C=\lbrace (x,\lambda)\in B^0\times\mathbb{C}\colon Tx=\lambda x\rbrace$ with the product topology which, being a closed subset of $B^0\times\mathbb{C}$, is σ-compact. Defining the continuous map $\pi\colon C\to\mathbb{C}$ by $\pi(x,\lambda)=\lambda$, then, $\sigma_p(T)=\pi(C)$ is σ-compact and, hence, is an Fσ set.
Now, let's move on to constructing $T$ with norm $\Vert T\Vert=R$ and point spectrum a given bounded Fσ set $K\subseteq B_R(0)$. The general case follows from the case where $K$ is closed, since if we write $K=\bigcup_{n=1}^\infty K_n$ for closed $K_n$ then, choosing operators $T_n$ with $\Vert T_n\Vert=R$ and $\sigma_p(T_n)=K_n$, the direct sum $T=\oplus_n T_n$ is an operator on $\oplus_n\ell^2\cong\ell^2$ with $\Vert T\Vert=R$ and point spectrum $K$ as required.
So, suppose that $K\subseteq B_R(0)$ is closed. By scaling, we can assume that $R > 1$ and that $K\subseteq B_1(0)$. I'll construct a linear operator $T$ on the Hilbert space $H=\ell^2(\mathbb{N}^2)\cong\ell^2$ (I'm using $\mathbb{N}=\lbrace0,1,2,\ldots\rbrace$). Recall that $H$ consists of the functions $f\colon\mathbb{N}^2\to\mathbb{C}$ with finite norm $\Vert f\Vert^2=\sum_{m,n}\vert f(m,n)\vert^2$ and inner product $\langle f,g\rangle=\sum_{m,n}\overline{f(m,n)}g(m,n)$. Now, choose sequences of positive real numbers $r_i,\epsilon_i$ and complex $z_i$ ($i=1,2,\ldots$) such that $\bar B_{r_i}(z_i)$ is disjoint from $K$ and such that $K=\bar B_1(0)\setminus\bigcup_iB_{r_i}(z_i)$. It doesn't matter exactly what values $\epsilon_i$ take at the moment, I'll choose these in more detail in a moment. Define $T\colon H\to H$ by $$ Tf(m,n)=\begin{cases} f(m,n+1),&\textrm{if }m=0,\cr r_mf(m,n-1)+z_mf(m,n),&\textrm{if }m,n > 0,\cr \epsilon_m r_m f(0,0)+z_mf(m,n),&\textrm{if }m > 0, n=0. \end{cases} $$ The idea is that $T$ behaves as a left-shift (on $m=0$), so that the point spectrum is contained in $\bar B_1(0)$, joined to a set of right-shifts (on each $m > 0$) to remove the balls $\bar B_{r_i}(z_i)$ from the point spectrum. It can be seen that $T$ has norm $$ \Vert T\Vert\le\max_m\left(1,\vert z_m\vert+r_m\right)+\left(\sum_m\epsilon_m^2r_m^2\right)^{1/2}. $$ This satisfies $\Vert T\Vert\le R$ so long as we choose $\bar B_{r_i}(z_i)\subseteq B_R(0)$ and choose $\epsilon_i$ so that $\sum_i\epsilon_i^2r_i^2$ is small enough. And (if you like), by scaling $\epsilon_i$ up, we can ensure that $\Vert T\Vert=R$.
Now, a complex number $\lambda$ is in the point spectrum $\sigma_p(T)$ if and only if there is a nonzero $f\in H$ such that $Tf=\lambda f$. Plugging in $Tf$ from the definition above and solving the linear equations shows that, up to a scaling factor, $f$ must be given by $$ f(m,n)=\begin{cases} \lambda^n,&\textrm{if }m=0,\cr \epsilon_m\left(r_m/(\lambda-z_m)\right)^{n+1},&\textrm{if }m > 0. \end{cases} $$ If $\vert\lambda\vert\ge1$ then $\sum_n\vert f(0,n)\vert^2=\infty$, so we must have $\vert\lambda\vert < 1$. Similarly, if $\lambda\in\bar B_{r_i}(z_i)$ then $\sum_n\vert f(i,n)\vert^2=\infty$. So, in order that $f\in H$ we must have $\lambda\in K$, in which case, $$ \Vert f\Vert^2=\frac{1}{1-\vert\lambda\vert^2}+\sum_{m=1}^\infty\frac{\epsilon_m^2}{\vert\lambda-z_m\vert^2/r_m^2-1}. $$ Now, as $\bar B_{r_i}(z_i)$ is disjoint from $K$, the term $(\vert\lambda-z_i\vert^2/r_i^2-1)^{-1}$ is a continuous positive function of $\lambda\in K$, so is bounded above by some $d_i^2 > 0$. Then, $$ \Vert f\Vert^2\le\frac{1}{1-\vert\lambda\vert^2}+\sum_{m=1}^\infty\epsilon_m^2d_m^2. $$ So long as we choose $\epsilon_i$ so that $\sum_m\epsilon_m^2d_m^2 < \infty$, this implies that $f\in H$ for all $\lambda\in K$, so $\sigma_p(T)=K$.
Note: The operator $T$ I constructed here does not give a positive answer to Q2. For any $\omega\in S^1$ and $\epsilon > 0$, setting $f_\epsilon(m,n)=1_{\lbrace m=0\rbrace}\epsilon\omega^n(1+\epsilon)^{-n-1}$ gives $\Vert f_\epsilon\Vert=1$ and $\Vert(T-\omega)f_\epsilon\Vert\to0$ as $\epsilon\to0$, so $\omega\in\sigma(T)$. Therefore, $S^1\subseteq\sigma(T)$.