Answer
see the proof below.
Work Step by Step
For any $p(x)=a_0+a_1x+...+a_nx^n, q(x)=b_0+b_1x+...+b_nx^n, r(x)=c_0+c_1x+...+c_nx^n \in P_n , k \in {R}$
(1) $\langle p,q\rangle=a_0^2+a_1^2+...+a_n^n>0$ and $\langle p,p\rangle=a_0^2+a_1^2+...+a_n^2=0$ if and only if $a_0=0$, $a_1=0$, ...,$a_n=0$.
(2) \begin{align*}
\langle p,q\rangle&= a_0b_0+a_1b_1+...+a_nb_n\\
&=b_0 a_0+ b_1a_1+...+b_na_n\\
&=\langle q,p\rangle.
\end{align*}
(3) \begin{align*}
\langle p,q\rangle&= ka_0b_0+ka_1b_1+...+ka_nb_n\\
&=k(a_0b_0+a_1b_1+...+a_nb_n)\\
&=k\langle p,q\rangle.
\end{align*}
(4)
\begin{aligned}
\langle p+q, r\rangle &=\left\langle\left(a_{0}+b_{0}\right)+\left(a_{1}+b_{1}\right) x+...+\left(a_{n}+b_{n}\right) x^{n}, c_{0}+c_{1} x+...+c_{n} x^{n}\right\rangle \\ &=\left(a_{0}+b_{0}\right) c_{0}+ \left(a_{1}+b_{1}\right) c_{1}+...+\left(a_{n}+b_{n}\right) c_{n} \\
&=\left(a_{0} c_{0}+2 a_{1} c_{1}+...+a_{n} c_{n}\right)+\left(b_{0} c_{0}+ b_{1} c_{1}+...+b_{n} c_{n}\right) \\ &=\langle p, r\rangle+\langle q, r\rangle .
\end{aligned}
