Route to go:
First show that $P\left(1\right)$ and $P\left(2\right)$are
both true.
Setting $B:=A_{1}\cup\cdots\cup A_{k}$ by applying $P\left(2\right)$
we find:
$\tag1 N\left(A_{1}\cup\cdots\cup A_{k}\cup A_{k+1}\right)=N\left(B\cup A_{k+1}\right)=N\left(B\right)+N\left(A_{k+1}\right)-N\left(B\cap A_{k+1}\right)$
Under assumption that $P(k)$ is true find expressions for: $$N\left(B\right)=N\left(A_{1}\cup\cdots\cup A_{k}\right)$$
and for $$N\left(B\cap A_{k+1}\right)=N\left(\left(A_{1}\cap A_{k+1}\right)\cup\cdots\cup\left(A_{k}\cap A_{k+1}\right)\right)$$
Substitute these expressions in (1).
edit:
$N\left(B\right)=\sum_{i=1}^{k}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\right)$
$N\left(B\cap A_{k+1}\right)=\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)-\cdots+\left(-1\right)^{k+1}N\left(\left(A_{1}\cap A_{k+1}\right)\cap\cdots\cap\left(A_{k}\cap A_{k+1}\right)\right)=\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)-\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
Substitution of this in the (1) gives the following RHS:
$\sum_{i=1}^{k}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\right)+N\left(A_{k+1}\right)-\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)+\cdots+\left(-1\right)^{k+2}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
After a rearrangement of the terms we find:
$N\left(A_{1}\cup\cdots\cup A_{k}\cup A_{k+1}\right)=\sum_{i=1}^{k+1}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k+1}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+2}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
wich is exactly statement $P\left(k+1\right)$.
Best Answer
For the first part, since $x^k \leq x^n$ for $k \leq n$ (whenever $x \geq 1$), the polynomial is at most $$(|a_n| + \cdots + |a_0|)x^n.$$
For the second part, the following facts are useful:
For the third part, the following facts are useful: