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 $u=(11,4)$ and $v=(8,-3)$, then we have
(a) since $u\cdot v=88-12=76\neq 0$, 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
$$a(11,4)+b(8,-3)=(0,0)$$
then we have the system
$$11a+8b=0, \quad 4a-3b=0.$$
The coefficient matrix has non zero determinant and hence the 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 $R^2$.
