Answer
(a) the set is not orthogonal.
(b) the set is not orthonormal.
(c) the set is a basis for $R^2$.
Work Step by Step
Let $u=(-4,6)$ and $v=(5,0)$, then we have
(a) since $u\cdot v=-20+0=-20$, then the set is not orthogonal.
(b) since the set is not orthogonal then it is not orthonormal.
(c) to check if it is a basis for $R^2$, consider the combination
$$a(-4,6)+b(5,0)=(0,0)$$
then we have the system
$$-4a+5b=0, \quad 6a=0.$$
It is clear that the above system has the solution $a=0, b=0$. Then the set is linearly independent and $R^2$ has dimension $2$, then it is a basis for $R^2$.
