Deciding on decidability of a problem and reducing it to halting problem if not decidable
The first language is semidecidable while the second one is not semidecidable.

Claim 1: $L_1 = \{ w\#x \hspace{1mm} |  \hspace{1mm} w,x \in \{0, 1\}^* \text{ where } M_w \text{ with input } x \text{ visits each of its states at least once} \}$ is semidecidable.
Proof: On input $w\#x$ Simply simulate $M_w$ on $x$ and count the number of visited states. If the counter reaches $|Q|$ then ACCEPT. This shows $L_1$ is semidecidable/RE. 


Claim 2: $L_2 = \{ w\#x \hspace{1mm} |  \hspace{1mm} w,x \in \{0, 1\}^* \text{ where } M_w \text{ with input } x \text{ does not visit at least one of its states} \}$ is not semidecidable.
Proof: Let $L_2$ be semidecidable. Then since $\overline{L}_1 = L_2$, i.e., $L_1$ and $L_2$ are complement of each other, $L_1$ (and so $L_2$) is decidable. In this case we can reduce the Halting problem to $L_1$ as following. Let $M$ be a Turing machine deciding $L_1$. We construct a Turing machine that decides the Halting problem. On input $w\#x$ we modify $M_w$ so that when $M$ halts, it first traverses all other states and then halts. We could do it by changing the ACCEPT state to a normal state, introducing a new ACCEPT state, and adding the following new transitions $$\delta(q_1, a) = q_2, \dots, \delta(q_n, a) = q_{new-accept}$$ where $a$ is a new alphabet symbol and when the modified $M_w$ is at the state $q_{old-accept}$ it first writes $a$ on the tape and enters the state $q_1$ (without moving the head, i.e, the head always reads $a$). Similarly for the the REJECT state. Thus $M_w(x)$ halts if and only if $M_w$ visits all states, which is decidable by $M$.