Sorry if the title doesn't completely explain the question but I couldn't find a way to fit it all in there. I am having some trouble with the following question:
In particular, I do not know how to prove or disprove these two, and I've have been unable to find counterexamples. How would you guys go about proving these two?
In the case of DFAs it's true; but for NFAs it's not in general true.
For DFAs Given $D$ and $D'$, that $L(D) = \overline {L(D')}$ follows easily from the definitions of the objects involved. In a little more detail, though:
The transition function $\delta$ has domain $Q \times \Sigma$, but it's easily extended to a function $\delta^{*}\colon Q \times \Sigma^{*} \to Q$ defined not just for letters but for all strings. $\delta^{*}(q, \sigma)$ is the state that $D$ will be in on input $\sigma$. $\delta^{*}$ is defined by induction on the length of strings $\sigma$. For $q \in Q, c \in \Sigma$: $$ \begin{align} \delta^{*}(q, \varepsilon) &= q \\ \delta^{*}(q, \sigma c) &= \delta(\delta^{*}(q, \sigma), c) \\ \end{align} $$ It's easy to see that a string $\sigma \in L(D)$ iff (if and only if) $\delta^{*}(s, \sigma) \in F$, where $s$ is the start state and $F$ the set of accepting states.
Given the definition of $D'$, its transition function is the same as that for $D$, so its extended transition function $\delta^{*}$ is the same too. A string $\sigma \in L(D')$ iff $D'$ accepts $\sigma$, iff $\delta^{*}(s, \sigma) \in Q \setminus F$, iff $\delta^{*}(s, \sigma) \notin F$, iff $\sigma \notin L(D)$. That is, the languages of $D, D'$ are complements of each other.
For NFAs, however, here's a counterexample $N = (Q, \Sigma, \Delta, s, F)$: $Q = \{q_0, q_1, q_2\}$, $\Sigma = \{a\}$, $s = q_0$, $F = \{q_1\}$; the transitions $\Delta$ are as follows: $$ (q_0, a) \to q_1 \\ (q_0, a) \to q_2 \\ (q_1, a) \to q_2 \\ (q_2, a) \to q_2 \\ $$ Thus, $L(N) = \{a\}$. In $N$, there are 2 paths on input $a$: one to state $q_1$ (accepting) and another to $q_2$ (non-accepting). In state $q_2$, any further letters in an input string remain in that non-accepting state, so strings longer than 1 are not accepted; in state $q_1$, any further letters in an input string transition on the first letter to $q_2$ then remain there, so again strings longer than 1 aren't accepted. The empty string is not accepted because $q_0$ isn't an accepting state.
If we obtain $N'$ from $N$ by switching the accepting states to the complement $\{q_0, q_2\}$, then $L(N') = \Sigma^{*}$, which also contains the string $a$.