Answer
see the proof below.
Work Step by Step
For any $p(x)=a_0+a_1x+a_1x^2, q(x)=b_0+b_1x+b_2x^2, r(x)=c_0+c_1x+c_2x^2 \in P_2 , k \in {R}$
(1) $\langle p,q\rangle=a_0^2+2a_1^2+a_3^2>0$ and $\langle p,p\rangle=a_0^2+2a_1^2+a_3^2=0$ if and only if $a_0=0$, $a_1=0$, $a_2=0$.
(2) \begin{align*}
\langle p,q\rangle&= a_0b_0+2a_1b_1+a_2b_2\\
&=b_0 a_0+2b_1a_1+b_2a_2\\
&=\langle q,p\rangle.
\end{align*}
(3) \begin{align*}
\langle p,q\rangle&= ka_0b_0+2ka_1b_1+ka_2b_2\\
&=k(a_0b_0+2a_1b_1+a_2b_2)\\
&=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_{2}+b_{2}\right) x^{2}, c_{0}+c_{1} x+c_{2} x^{2}\right\rangle \\ &=\left(a_{0}+b_{0}\right) c_{0}+2\left(a_{1}+b_{1}\right) c_{1}+\left(a_{2}+b_{2}\right) c_{2} \\ &=\left(a_{0} c_{0}+2 a_{1} c_{1}+a_{2} c_{2}\right)+\left(b_{0} c_{0}+2 b_{1} c_{1}+b_{2} c_{2}\right) \\ &=\langle p, r\rangle+\langle q, r\rangle .
\end{aligned}
