$$ H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 +V_0 x^3 $$ Find $\frac{d\langle x\rangle}{dt}$, $\frac{d\langle p\rangle}{dt}$ and $\frac{d\langle H\rangle}{dt}$.
If this is the hamiltonian given then this should be impossible, as there is no time variable. Am i right? I have trouble evaluating all three.
Assuming that we work in the Schrödinger picture: the time evolution of the system is encoded in its wave function, as determined by the time-dependent Schrödinger equation: $$i\hbar\partial_t\Psi(x, t) = \hat{H}\Psi(x,t).$$ The time-dependent expectation values are then given by $$\langle x(t)\rangle = \langle \Psi(t)|\hat{x}|\Psi(t) = \int dx\Psi(x,t)^*\hat{x}\Psi(x,t), etc.$$ Generally speaking, these are time-dependent and have non-zero time derivatives.
However, if the system is in a stationary state, i.e. $$\hat{H}\psi(x) = E\psi(x),$$ then the time dependence of the wave function is trivial $$\Psi(x,t)=\psi(x)e^{-iEt/\hbar},$$ and the averages in question are time-independent.
Finally, let us note that $\frac{d\langle x(t)\rangle}{dt}$ is not the same as $\left\langle\frac{d x(t)}{dt}\right\rangle$ - the former is the derivative of an expectation value, whereas the latter is the expectation of the derivative, where the operator of the derivative is defined by $$\hat{\dot{x}} = \frac{d \hat{x}(t)}{dt} = \frac{1}{i\hbar}\left[\hat{x}, \hat{H}\right].$$ Note that, as we are still in the Schrödinger picture, this is definition of the operator of the derivative rather than a time derivative of the operator. In the Heisenberg and interaction pictures the latter is given by the same equation, but with time-dependent operators.
