Problem understanding DFA & NFA equivalence in Theory of Computation
Motivation. In this construction, the intent is to make a DFA equivalent to a given NFA. In the original NFA, look at what happens when we're in state 2 and we see input $a$: we could either stay in state 2 or make a transition to state 3. If we want to eliminate nondeterminism, we could say that from state 2 looking at input $a$ we move to the single new state $\{2,3\}$, meaning that in this case, we're either in state 2 or state 3. In other words, in the constructed DFA we'll have the transition $\delta'(\{2\}, a)=\{2, 3\}$. The important thing to notice is that this transition in the new FA is deterministic: from the state $\{2\}$ on reading $a$ we move to the single state denoted by $\{2, 3\}$. Guided by this convention, we will build the DFA by defining the states to be all possible sets over 1, 2, 3, namely 
$$
\emptyset, \{1\}, \{2\},\{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}
$$

Final States. Let's start with an example: what does the NFA do on input $baa$? We'll have, in order


In state 1, looking at $b$. Go to state 2.
In state 2, looking at $a$. Go to state 2 or state 3.
Looking at the last input, $a$ we could be in either state 2 or state 3. If we're in state 2, go to state 2 or state 3 as before. If we're in state 3, go to state 1. 


In other words, at this stage, after having read $baa$ we could find ourself in states 2, 3, or 1.

In the DFA, we have, equivalently, the transitions $\delta'$ defined as


$\delta'(\{1\}, b) = \{2\}$.
$\delta'(\{2\},a) = \{2, 3\}$.
$\delta'(\{2, 3\}, a) = \delta'(\{2\},a)\cup\delta'(\{3\},a) = \{2, 3\}\cup\{1\}=\{1, 2, 3\}$.


Now the question is, should state $\{1, 2, 3\}$ be considered as a final state? Sure, since the NFA, reading $baa$ has at lest one possible collection of transitions that leads to the accepting state 1. In a similar way, any state of the DFA containing 1 will be a final state, so in the DFA there will be four final states: 
$$
\{1\}, \{1, 2\}, \{1, 3\}, \{1, 2, 3\}
$$
Now it turns out that in the DFA the state $\{1,2\}$ is never used, but that doesn't negate the fact that it's a final state (since it contains 1). All it means is that there is no transition that takes you there. It's what we call an unreachable state. This, I hope, answers your first question.

Start State. This answer to your first question has little to do with the answer to your second question. The answer for your second question is: just as we did to determine the final states of $D$, we also need to know what the start state of $D$ is, and in this case it will be the set of states containing the start state of $N_4$, along with any other states that can be reached from that start state via $\epsilon$-moves, namely $\{1, 3\}$ (which of course will also be a final state, though that's not important here).