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 $\ell^2$.
Theorem: A set $K\subseteq\mathbb{C}$ is equal to the point spectrum of a bounded linear operator $T\colon \ell^2\to \ell^2$ if and only if $K$ is bounded and is the union of countably many closed sets (an Fσ set).
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, $S^1\subseteq\sigma(T)$.
Best Answer
Yes. If $E$ is empty, let $X=\{0\}$. (By the Gelfand-Mazur theorem, this is the only possibility if $\mathbb{F}=\mathbb{C}$.)
If $E$ is not empty, we may assume without loss of generality that $0\in E$. Now take a Banach space $Y$ with a bounded operator $T:Y\to Y$ that is quasinilpotent (i.e., its spectrum is $\{0\}$) but not nilpotent. Let $B(Y)$ be the algebra of bounded operators on $Y$ and let $X\subseteq B(Y)$ be the subalgebra generated by $T$ and the elements $(T-\lambda I)^{-1}$ for all $\lambda\in \mathbb{F}\setminus E$ (here we use the assumption that $0\in E$, so that all these inverses must exist). Note that the assumption that $T$ is not nilpotent means that that no nontrivial polynomial in $T$ is zero, so $X$ is isomorphic as an $\mathbb{F}$-algebra to the subalgebra of the field of rational functions $\mathbb{F}(x)$ generated by $x$ and $(x-\lambda)^{-1}$ for $\lambda\in\mathbb{F}\setminus E$, by mapping $T$ to $x$. In particular, $T-\lambda I$ is not invertible in $X$ for any $\lambda\in E$.
Now let $A:X\to X$ be the operator given by multiplication by $T$. Then $A$ is bounded since $X$ is a normed algebra. Also, $A-\lambda I$ is invertible for any $\lambda\in\mathbb{F}\setminus E$ (the inverse is just multiplication by $(T-\lambda I)^{-1}$) but not for any $\lambda\in E$. Thus, the spectrum of $A$ is $E$.