Answer
see the proof below.
Work Step by Step
For any $(u_1,u_2), (v_1, v_2 ), (w_1, w_2) \in {R}^2 , k \in {R}$
(1) $\langle (u_1,u_2), (u_1, u_2 )\rangle=2 u_1^2+ 2u_1u_2+2u_2^2=u_1^2+u_2^2+(u_1+u_2)^2>0$ and $\langle (u_1,u_2), (u_1, u_2 )\rangle=u_1^2+u_2^2+(u_1+u_2)^2=0$ if and only if $u_1=0$, $u_2=0$.
(2) \begin{align*}
\langle (u_1,u_2), (v_1, v_2 )\rangle&= 2u_1v_1+ u_1v_2+ u_2v_1+2u_2v_2\\
&=2v_1u_1+ v_2u_1+ v_1u_2+2v_2u_2\\
&=\langle (v_1, v_2 ),(u_1,u_2)\rangle.
\end{align*}
(3) \begin{align*}
\langle (ku_1,ku_2), (v_1, v_2 )\rangle&=2ku_1v_1+ ku_1v_2+ ku_2v_1+2k u_2v_2\\
&=k(2u_1v_1+ u_1v_2+ u_2v_1+2u_2v_2)\\
&=k\langle (u_1,u_2), (v_1, v_2 )\rangle.
\end{align*}
(4) $$
\begin{aligned}\left\langle\left(u_{1}, u_{2}\right)+\left(v_{1}, v_{2}\right),\left(w_{1}, w_{2}\right)\right\rangle &=\left\langle\left(u_{1}+v_{1}, u_{2}+v_{2}\right),\left(w_{1}, w_{2}\right)\right\rangle \\ &= 2\left(u_{1}+v_{1}\right) w_{1}+\left(u_{1}+v_{1}\right) w_{2}+\left(u_{2}+v_{2}\right) w_{1}+2\left(u_{2}+v_{2}\right) w_{2} \\ &=\left(2 u_{1} w_{1}+u_{1} w_{2}+u_{2} w_{1}+2 u_{2} w_{2}\right)+\left(2 v_{1} w_{1}+v_{1} w_{2}+v_{2} w_{1}+2 v_{2} w_{2}\right) \\ &=\left\langle\left(u_{1}, u_{2}\right),\left(w_{1}, w_{2}\right)\right\rangle+\left\langle\left(v_{1}, v_{2}\right),\left(w_{1}, w_{2}\right)\right\rangle \end{aligned}
$$
