It depends on the multiplication algorithm used. Let $n$ be the bit size of the finite field used, and let $O(n^k)$ be the complexity of the multiplication algorithm used (e.g. $k=2$ for schoolbook multiplication, or $k=1.585$ for Karatsuba multiplication). Then:
- Point addition: $O(n^k)$ (a constant number of multiplications)
- Scalar multiplication: $O(n^{k+1})$
- Selecting random point: depends on the method, but usually the same as scalar multiplication.
As pointed out by poncho, the point addition (and the scalar multiplication) can require a multiplication inverse, whose complexity again depends on the algorithm and may be larger than the complexity of multiplication. For example, binary euclidean takes $O(n^2)$, Fermat's little theorem takes $O(n^{k+1})$.