Necessary conditions for construction of $n$ to $p(n)$ pseudorandom generators from $n$ to $n+1$ generators?
It is not necessary, that $G_1$ is a PRG.

Let $G_2: \{0,1\}^{n-1} \rightarrow \{0,1\}^n$ be a PRG, define $G_1: \{0,1\}^n \rightarrow \{0,1\}^{n+1}$ as
\begin{align*}
G_1(s_1||\ldots||s_n) := 1||G_2(s_2||\ldots||s_n).
\end{align*}
and consider the distinguisher $\mathcal{D}_1$, which returns the first bit $w_1$, when given the $n+1$ bit string $w := w_1||\ldots||w_{n+1}$ as input.

For $s_1 \leftarrow \{0,1\}^n$ and $r_1 \leftarrow \{0,1\}^{n+1}$ we have
\begin{align*}
\Big|\text{Pr}[\mathcal{D}_1(G_1(s_1))=1] - \text{Pr}[\mathcal{D}_1(r_1)=1]\Big| = \Big|1 - \frac{1}{2}\Big| = \frac{1}{2} &gt; \mathsf{negl}(n),
\end{align*}
which implies that $G_1$ is not a PRG. 

Nevertheless, we can still prove, that $G$ is a PRG, by using a slight modification of Katz' and Lindell's proof for theorem 6.20 in "Introduction to Modern Cryptography". (I assume you are familiar with it, since you are using the exact same notation.)

We construct a distinguisher $\mathcal{D}_2$ for $G_2$ using another distinguisher $\mathcal{D}$ for $G$, with
\begin{align*}
\Big|\text{Pr}[\mathcal{D}(G(s))=1] - \text{Pr}[\mathcal{D}(r)=1]\Big| =: \epsilon(n).
\end{align*}

Let $\mathcal{D}_2$ behave as follows, when given a $w := w_1 || \ldots || w_n \in \{0,1\}^n$ as input.


$i \leftarrow \{0, \ldots, p(n) - 1\}$
$r \leftarrow \{0,1\}^i$
$p := G(1||w)$
Construct a bit-string $w'$ as the concatenation of $r$, $w_{n}$ and the first $p(n) - i - 1$ bits of $p$.
Output $\mathcal{D}(w')$


You can easiliy verifiy the following two statements:


If $i = 0$ and $w = G_2(s_2)$ for $s_2 \in \{0,1\}^{n-1}$, then $w' = G(s)$ for $s = b||s_2$ and $b \in \{0,1\}$.
If $i = p(n) -1$ and $w = r_2 \leftarrow \{0,1\}^n$, then
$w' \leftarrow \{0,1\}^{p(n)}$.


Using the same argument as Katz and Lindell, we obtain
\begin{align*}
\mathsf{negl}(n) \geq \frac{\epsilon(n)}{p(n)} &amp;&amp; \Longrightarrow &amp;&amp; \mathsf{negl}(n) \geq \epsilon(n),
\end{align*}
which means, that $G$ is a PRG.

Similar arguments apply to your other two questions.