Yes, there are. For (a) note, that the spectrum is closed. And the set you gave, namely $\Pi := \{\pi_n \mid n \in \mathbb N\}$ need be closed. As the spectrum is closed, we obtain $\overline\Pi \subseteq \sigma(M_\pi)$, we will now show the other inclusion: Let $\lambda \not\in \overline\Pi$, then $\mathrm{dist}(\lambda, \Pi)> 0$, define $\psi_n := (\pi_n - \lambda)^{-1}$, then $|\psi_n| \le \frac 1{\mathrm{dist}(\lambda,\Pi)}$, hence $\psi := (\psi_n)$ is bounded and $M_{\psi}$ is inverse to $M_\pi - \lambda$. So $\sigma(M_\pi) = \overline\Pi$.
For (b) note, that if $M_\pi$ is compact, by $M_\pi e_n = \pi_n e_n$ we need $(\pi_n e_n)$ to have a convergent subsequence, with 0 as only possible limit (as $\ell^2$-convergence implies coordinatewise convergene). Applying this argument to each subsequence, we obtain $\pi_n \to 0$ if $M_\pi$ is compact. On the other hand, suppose $\pi_n \to 0$. Consider for $n \in \mathbb N$ the sequence $ \pi^{(n)} := (\pi_1, \ldots, \pi_n, 0, \ldots) $and the corresponding multiplication operator $M_{\pi^{(n)}}$. $M_{\pi^{(n)}}$, having finite dimensional range, is compact, and for $x \in \ell^2$ we have \begin{align*}\def\norm#1{\left\|#1\right\|} \norm{M_\pi x - M_{\pi^{(n)}}x} &= \left(\sum_{k=n+1}^\infty |\pi_kx_k|^2\right)^{1/2}\\ &\le \sup_{k \ge n+1}|\pi_k| \cdot \left(\sum_{k=n+1}^\infty |x_k|^2\right)^{1/2}\\ &\le \sup_{k\ge n+1} |\pi_k| \cdot \|x\| \end{align*} Hence $M_{\pi^{(n)}} \to M_\pi$ in $L(\ell^2)$, and $M_\pi$, being the norm-limit of compact operators is compact. So $M_\pi$ is compact iff $\pi$ converges to zero.