If $A$ is an abstract $C^*$ algebra, then a $A$ is isometrically isomorphic to a subalgebra of bounded operators $B(H)$ for some Hilbert space $H$. We can restrict the weak operator toplogy and the strong operator topology of $B(H)$ to $A$.
Are these definitions independent of $H$? And what about the case of $A$ being an abstract Von Neummann Algebra (a $C^*$ algebra who has a predual)?
No, the weak and strong operator topologies depend strongly on the chosen representation.
For example, if you represent $C([0,1])$ on $L^2([0,1])$ by multiplication operators, then the functions $$f_n(x) = x^n$$ converge strongly to zero. However, if you take the direct sume of that representation with the one dimensional representation corresponding to the character $$f\mapsto f(1),$$ then the $f_n$ do not converge strongly to any $f$ in $C([0,1])$.