This question has been bugging me for a while but, now, I can finally give a positive answer to question 2. In fact, we can say exactly which sets occur as the spectrum and point spectrum of a linear operator on $\ell^2$.
Theorem. Let $K_1,K_2$ be subsets of $\mathbb{C}$. Then, there exists a bounded linear operator $T\colon\ell^2\to\ell^2$ with $\sigma_p(T)=K_1$ and $\sigma(T)=K_2$ if and only if (i) $K_2$ is a nonempty closed and bounded set and (ii) $K_1\subseteq K_2$ is an Fσ set (a countable union of closed sets).
First, these conditions on $K_1,K_2$ are necessary. By standard properties of bounded operators, $\sigma(T)=K_2$ must be closed, bounded and nonempty. Also, the point spectrum must be contained in the spectrum and, as shown in my other answer to this question (the answer to Q1), is an Fσ set.
Let's start with the case where $K=K_1=K_2$ is a nonempty closed and bounded set.
I'll construct an example with the following properties.
$H$ is a separable Hilbert space and $T\colon H\to H$ is a bounded operator whose point spectrum includes the closed ball $\bar B_{1/2}(0)$. For any closed nonempty $K\subseteq\bar B_{1/2}(0)$, let $H_K$ be the closed subspace of $H$ generated by the eigenvectors of $T$ with eigenvalues in $K$ and let $T_K\colon H_K\to H_K$ be the restriction of $T$ to $H_K$. Then, $\sigma_p(T_K)=\sigma(T_K)=K$.
So, we can find examples with spectrum $\sigma(T)=\sigma_p(T)=K$ by restricting to subspaces $H_K$ of a single given separable Hilbert space $H$. Then, $H_K$ is a separable Hilbert space so is either isomorphic to $\ell^2$ or is finite dimensional (in which case the direct sum of a countable set of copies of $H_K$ is isomorphic to $\ell^2$). We can then find examples for any closed bounded nonempty set $K$ by scaling.
First, I'll set up a bit of notation. Choose a finite sequence $z_0,z_1,\ldots,z_{N-1}\in B_{1/2}(0)$ such that $\bigcup_{n=0}^{N-1}\bar B_{1/4}(z_n)\supseteq\bar B_{1/2}(0)$. Let $[N]=\lbrace0,1,\ldots,N-1\rbrace$ and define $[N]^\ast$ to be the disjoint union of $[N]^n$ over $n\in\mathbb{N}$ (I'm using $\mathbb{N}=\lbrace0,1,\ldots\rbrace$). Note that $[N]^\ast$ is countable. For each $a=(a_0,\ldots,a_{n-1})\in[N]^n$ set $z_a=z_{a_0}+2^{-1}z_{a_1}+\cdots+2^{-(n-1)}z_{a_{n-1}}$ and $r_a=2^{-n}$. Also set $a\cdot k=(a_0,\ldots,a_{n-1},k)\in[N]^{n+1}$ for each $k\in[N]$. By construction we have,
$$
\begin{align}
&B_{r_{()}}(z_{()})=B_1(0),\cr
&\bar B_{\frac{r_a}{2}}(z_a)\subseteq\bigcup_{k\in[N]}\bar B_{\frac{r_{a\cdot k}}{2}}(z_{a\cdot k})
\end{align}
$$
Now define a separable Hilbert space $V=\bigoplus_{a\in[N]^\ast}\ell^2$. Elements $f\in H$ are maps $f\colon[N]^\ast\to\ell^2$ with finite norm $\Vert f\Vert^2=\sum_{a\in[N]^*}\Vert f(a)\Vert^2$, and inner product $\langle f,g\rangle=\sum_{a\in[N]^\ast}\langle f(a),g(a)\rangle$. Define the shift map $S\colon\ell^2\to\ell^2$ by $S f(n)=f(n+1)$. Then, $\Vert S^n\Vert=1$ for all $n\ge 0$ in which case, for all $\lambda\in\mathbb{C}$ with $\vert\lambda\vert > 1$,
$$
\begin{align}
&(\lambda-S)^{-1}=\sum_{n=0}^\infty\lambda^{-n-1}S^n,\cr
&\Vert(\lambda-S)^{-1}\Vert\le\sum_{n=0}^\infty\vert\lambda\vert^{-n-1}=(1-\vert\lambda\vert)^{-1}.
\end{align}
$$
So, $\sigma(S)\subseteq\bar B_1(0)$. Also, for $\vert\lambda\vert < 1$ set $x_{\lambda}=(1,\lambda,\lambda^2,\ldots)\in\ell^2$ so that $S x_\lambda=\lambda x_\lambda$. In fact, it is easily seen that every eigenector of $S$ with eigenvalue $\lambda$ is a multiple of $x_\lambda$. We also have $\Vert x_\lambda\Vert^2=(1-\vert\lambda\vert^2)^{-1}$.
Define the linear map $T\colon V\to V$ by
$$
Tf(a) = (r_aS+z_a)f(a).
$$
As $B_{r_a}(z_a)\subseteq B_1(0)$ this is a bounded map with norm $\Vert T\Vert\le1$. Note that $f\in V$ is an eigenvector of $T$ with eigenvalue $\lambda$ if and only if $f(a)$ is proportional to $x_{(\lambda-z_a)/r_a}$ whenever $\lambda\in B_{r_a}(z_a)$ and $f(a)=0$ otherwise.
Now, for any $\alpha,\beta\in\mathbb{C}$ with $\vert\alpha\vert+\vert\beta\vert < 1$ define $U_{\alpha,\beta}\colon\ell^2\to\ell^2$ by
$$
U_{\alpha,\beta}f(n)=\sum_{k=0}^n\binom{n}{k}\alpha^{n-k}\beta^kf(n-k).
$$
This is a bounded linear operator with norm
$$
\begin{align}
\Vert U_{\alpha,\beta}\Vert&\le\sum_{k=0}^\infty\sup_{n\ge k}\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert\le\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert\cr
&=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert=\sum_{n=0}^\infty(\vert\alpha\vert+\vert\beta\vert)^n\cr
&=\left(1-\vert\alpha\vert-\vert\beta\vert\right)^{-1}.
\end{align}
$$
It can be seen that this moves the eigenvectors of $S$ according to $U_{\alpha,\beta}x_\lambda=x_{\alpha\lambda+\beta}$ and, also, satisfies the commutation relation $SU_{\alpha,\beta}=U_{\alpha,\beta}(\alpha S+\beta)$ (you just need to check that this holds on the eigenvectors $x_\lambda$). Define the map $U\colon V\to V$ by
$$
Uf(a)=\sum_{k\in[N]}U_{1/2,z_k}f(a\cdot k).
$$
This has finite norm,
$$
\Vert U\Vert\le\sum_{k\in[N]}\Vert U_{1/2,z_k}\Vert < \infty.
$$
Furthermore, from the commutation relations for $S,U_{\alpha,\beta}$, it can be seen that $T$ and $U$ commute. Fixing any $c\in\mathbb{C}$ with $0 < \vert c\vert < 1$, let $H$ be the set of $f\in V$ with $Uf=cf$, which is a closed subspace of $H$. Let $T_H\colon H\to H$ be the restriction of $T$ to $H$.
With the (rather lengthy!) setup done, we can now show that $T_H$ has the required properties. Choosing any $\lambda\in\bar B_{1/2}(0)$ we can show that there is an $f\in H$ with $Tf=\lambda f$, so that $\bar B_{1/2}(0)\subseteq\sigma_p(T_H)$. By construction, there is a sequence $a_0,a_1,a_2,\ldots\in[N]$ such that, setting $a_{(n)}=(a_0,a_1,\ldots,a_{n-1})\in[N]^n$ then $\lambda\in\bar B_{\frac{r_{a_{(n)}}}{2}}(z_{(n)})$ for each $n\in\mathbb{N}$. Define $f\in H$ by $f(a_{(n)})=c^nx_{(\lambda-z_{a_{(n)}})/r_{a_{(n)}}}$ and $f(b)=0$ for all $b\in[N]^\ast$ not equal to any of the $a_{(n)}$. We need to check that this does have finite norm, so is really in $H$. As $\lambda\in\bar B_{\frac{r_{a_{(n)}}}{2}}(z_{(n)})$, we have $\Vert f(a_{(n)})\Vert^2\le\vert c\vert^{2n}(1-1/4)^{-1}$. So,
$$
\Vert f\Vert^2\le\sum_n \frac43\vert c\vert^{2n}=\frac{4}{3(1-\vert c\vert^2)} < \infty.
$$
Also, $T f=\lambda f$ so $\sigma_p(T_H)\supseteq\bar B_{1/2}(0)$ as required.
Now choose any nonempty closed $K\subseteq\bar B_{1/2}(0)$, let $H_K$ be the closed subspace of $H$ generated by eigenvectors of $T_H$ and $T_K\colon H_K\to H_K$ be the restriction of $T$ to $H_K$. We automatically have $K\subseteq\sigma_p(T_K)\subseteq\sigma(T_K)$, so it just needs to be shown that $\sigma(T_K)\subseteq K$. That is, fixing $\lambda\in\mathbb{C}\setminus K$, we need to show that $\lambda-T_K$ is invertible. Let $d$ be the distance from $\lambda$ to the set $K$, and choose $n\ge0$ so that $d > 2^{1-n}$. Define the linear map $W\colon V\to V$ by
$$
Wf(a)=1_{\lbrace a\in[N]^m,m\ge n\rbrace}1_{\lbrace B_{r_a}(z_a)\cap K\not=\emptyset\rbrace}(\lambda-r_aS-z_a)^{-1}f(a).
$$
Note that if $B_{r_a}(z_a)\cap K\not=\emptyset$ and $a\in[N]^m$ for some $m\ge n$ then $z_a$ is within $r_a$ of $K$. So, $\vert\lambda-z_a\vert\ge d-r_a$ and, hence,
$$
\Vert(\lambda-r_aS-z_a)^{-1}\Vert\le r_a^{-1}(\vert\lambda-z_a\vert/r_a-1)^{-1}\le(d-2r_a)^{-1}\le(d-2^{1-n})^{-1}.
$$
So $W$ is a bounded operator with $\Vert W\Vert\le(d-2^{1-n})^{-1}$.
If $f\in H$ is an eigenvalue of $T$ with eigenvector $\mu\in K$ then we have $f(a)=0$ whenever $B_{r_a}(z_a)\cap K=\emptyset$. So, $(\lambda-T)Wf(a)=W(\lambda-T)f(a)=f(a)$ for all $a\in[N]^m$ with $m\ge n$. So, if we also have $f\in H_K$, then
$$
c^{-n}U^nW(\lambda-T)f=f.
$$
So, $L\Vert(\lambda-T)f\Vert\ge\Vert f\Vert$ for all $f\in H_K$, where I have set $L=\vert c\vert^{-n}\Vert U^n\Vert$. We can define $(\lambda-T_K)$ first by defining on the eigenvectors with eigenvalues $\mu\in K$ (which is just multiplication by $(\lambda-\mu)^{-1}$, so is well defined), noting that this is a linear map on the space generated by the eigenvalues with norm bounded by $L$ and, hence, has a unique continuous extension to $H_K$. So, $\lambda\not\in\sigma(T_K)$, showing that $\sigma(T_K)\subseteq K$ as required.
Now, consider the case where $K_1$ is an Fσ set and $K_2$ is its closure (assumed to be nonempty and bounded). Then, we can write $K_1=\bigcup_{n=1}^\infty K_{1,n}$ for nonempty closed sets $K_{1,n}$. By scaling, we suppose that $K_2\subseteq\bar B_{1/2}(0)$. Let $H$ be the separable Hilbert space and $T\colon H\to H$ be the bounded linear map described above. Then, $H_{K_{1,n}}$ is the closed subspace of $H$ generated by the eigenvectors of $T$ with eigenvalues in $K_{1,n}$ and $T_{K_{1,n}}\colon H_{K_{1,n}}$ is the restriction of $T$ to $H_{K_{1,n}}$. As explained above, $\sigma_p(T_{K_{1,n}})=\sigma(T_{K_{1,n}})=K_{1,n}$. Also, $\Vert T_{K_{1,n}}\Vert\le\Vert T\Vert$.
Set $\tilde H=\oplus_n H_{K_{1,n}}$ which is a separable Hilbert space, so is isomorphic to $\ell^2$. Define $\tilde T\colon\tilde H\to\tilde H$ by $\tilde T\oplus_n T_{K_{1,n}}$. This is a bounded linear map with $\Vert\tilde T\Vert\le\Vert T\Vert$ and point spectrum $\bigcup_n K_{1,n}=K_1$. It just needs to be shown that $\sigma(\tilde T)\subseteq K_2$ or, equivalently, for any $\lambda\in\mathbb{C}\setminus K_2$ that $\lambda-\tilde T$ is invertible. If its inverse exists, it must be of the form $(\lambda-\tilde T)^{-1}=\oplus_n(\lambda-T_{K_{1,n}})^{-1}$ which has norm $\sup_n\Vert(\lambda-T_{K_{1,n}})^{-1}\Vert$. It needs to be shown that this norm is finite. Note that, as $\lambda\not\in K_2$ then $\lambda-T_{K_2}$ is invertible, and $\lambda-T_{K_{1,n}}$ is just its restriction to $H_{K_{1,n}}$. So, $\Vert(\lambda-T_{K_{1,n}})^{-1}\Vert\le\Vert(\lambda-T_{K_2})^{-1}\Vert$ is bounded independently of $n$, as required.
Finally, consider the general case where $K_1$ is an Fσ set and $K_2\supseteq K_1$ is closed, bounded and nonempty. In the case where $K_1$ is empty then we need to find a bounded operator with empty point spectrum and given spectrum $K_2$. If $K_1$ is nonempty, then an operator of the form $T_1\oplus T_2$ on $\ell^2\oplus\ell^2$ will work, where $T_1$ has point spectrum $K_1$ and spectrum $K_2$. In either case, it reduces to finding a bounded linear operator $T$ with empty point spectrum and spectrum a given closed bounded and nonempty set $K$. Define the right-shift $S\colon\ell^2\to\ell^2$ by $Sf(n)=1_{\lbrace n\ge1\rbrace}f(n-1)/n$. Then, $S^kf(n)=1_{\lbrace n\ge k\rbrace}f(n-k)/(n(n-1)\cdots(n-k+1))$. So, $\Vert S^k\Vert=1/k!$. Therefore, for any $\lambda\in\mathbb{C}$ with $\lambda\not=0$ we have
$$
\begin{align}
&(\lambda-S)^{-1}=\sum_{n=0}^\infty\lambda^{-1-n}S^n\cr
&\Vert(\lambda-S)^{-1}\Vert\le\sum_{n=0}^{\infty}\vert\lambda\vert^{-1-n}/n!=\vert\lambda\vert^{-1}\exp(\vert\lambda\vert^{-1}).
\end{align}
$$
So, $S$ has spectrum $\lbrace0\rbrace$. It is easily seen to have no eigenvalues, so empty point spectrum. Now choose a sequence $z_1,z_2,\ldots$ dense in $K$ and set $T=\oplus_n(S+z_n)$. This has empty point spectrum and spectrum containing $z_n$, so $\sigma(T)\supseteq K$. On the other hand, if $\lambda\not\in K$ then, letting $d$ be the distance of $\lambda$ from $K$, the operators $(\lambda-s-z_n)^{-1}$ have norm bounded by $d^{-1}\exp(d^{-1})$. So, $(\lambda-T)^{-1}=\oplus_n(\lambda-S-z_n)^{-1}$ is a bounded linear operator, and $\lambda\not\in\sigma(T)$. Therefore, $\sigma(T)=K$.
