Answer
The set $\{(\frac{3}{5},\frac{4}{5}),(-\frac{4}{5},\frac{3}{5})\}$is orthogonal, not orthonormal and it is a basis for $R^2$.
Work Step by Step
given $\{(\frac{3}{5},\frac{4}{5}),(-\frac{4}{5},\frac{3}{5})\}$
let $v_1=(\frac{3}{5},\frac{4}{5}),v_2=(-\frac{4}{5},\frac{3}{5})$
(a)
since $v_1v_2=-\frac{12}{5}+\frac{12}{5}=0$
then, the set $\{(\frac{3}{5},\frac{4}{5}),(-\frac{4}{5},\frac{3}{5})\}$ is orthogonal.
(b)
scince
$\left\|{v}_{1}\right\|=\sqrt{v_{1} \cdot v_{1}}=\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{\frac{25}{25}}=1$
$\left\|{v}_{2}\right\|=\sqrt{v_{2} \cdot v_{2}}=\sqrt{\frac{16}{25}+\frac{9}{25}}=\sqrt{\frac{25}{25}}=1$
then, the set $\{(\frac{3}{5},\frac{4}{5}),(-\frac{4}{5},\frac{3}{5})\}$ is orthonormal.
(c)
by the corollary to Theorem 5.10, $\{(\frac{3}{5},\frac{4}{5}),(-\frac{4}{5},\frac{3}{5})\}$ is a basis for $R^2$.
