I know that expected value of velocity means $$\langle v\rangle=\frac{d\langle x\rangle}{dt}$$ and since I know how to calculate $\langle x\rangle$ namely $$\langle x\rangle=\int_{-\infty}^{\infty}x|\Psi|^2dx,$$ I can compute $\langle v\rangle$.
Now my question is what is meant by $\langle v^2\rangle$? Does it mean $\frac{d\langle x^2\rangle}{dt}$?
In the context of (non-relativistic) quantum mechanics there is no velocity operator. There is instead the momentum operator whose powers you want to calculate the expectation value of, on a state $|\alpha\rangle$.
Thus one has, by definition: $$ \langle \hat{p}^2\rangle = \langle \alpha|\hat{p}^2|\alpha\rangle $$ In order to explicitly calculate the above quantity one must know how the momentum operator acts on the state $|\alpha\rangle$. Usually you can express $|\alpha\rangle = \sum_{a}c_a|\phi_{a}\rangle$ with $|\phi_{a}\rangle$ being a complete set of eigenstates of the Hamiltonian, and maybe you know how the momentum operator acts on the energy eigenstates.
Otherwise, you can insert the identity operator in the bra-ket scalar product and re-write everything in terms of integrals and derivatives of the wave function, in case you have an explicit expression thereof.
No it is not the same as $\frac{d\langle x^2\rangle}{dt}$. This is because the derivative will become $2\langle x \rangle \frac{d\langle x\rangle}{dt}$ (I've basically used chain rule/ implicit differentiation). For this one, it'll be easier to work out $\langle p^2 \rangle$ first using the momentum operator :)
No. Suppose we have $\hat{H}$ and an observable $\hat{A}$ both independent of time, from the Schrodinger equation $\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$ we have
$$\frac{d}{dt} \langle \psi(t)| \hat{A} |\psi(t)\rangle = \frac{\text{i}}{\hbar}\langle \psi(t)| [\hat{H},\hat{A}] |\psi(t)\rangle $$
If $\hat{A}=\hat{X}$ then we recover your first identity. But if $\hat{A}=\hat{X^2}$, you can easily compute the commutator and realize it is not $\hat{V}^2$.
The mathematical definition of $\langle \hat{V}^2 \rangle$ is given on the first line provided $|\psi(t)\rangle$ is normalized, and can be computed, for example, in the momentum basis as the last two lines \begin{align*} \langle \hat{V}^2 \rangle &=\langle\psi(t)| \frac{\hat{P}^2 }{m^2} |\psi(t) \rangle \\ &=\int_{-\infty}^{\infty} dp \langle\psi(t)|p\rangle \langle p| \frac{\hat{P}^2 }{m^2}|\psi(t)\rangle\\ &=\int_{-\infty}^{\infty} dp |\langle p|\psi(t)\rangle|^2 \frac{p^2}{m^2} \end{align*}
The last expression tells you what it means: If you assume a large collection ($N \longrightarrow \infty$) of systems each one in the state $|\psi(t)\rangle$ and measure the momentum (or the velocity) you can get its distribution, $|\langle p|\psi(t)\rangle|^2$, which then allows you to get the average of squared velocity.
Additionally this average is related to the average kinetic energy since \begin{align*} \langle \hat{T}\rangle = \frac{1}{2}m \langle \psi(t)|\hat{V}^2 |\psi(t)\rangle \end{align*} and satisfies the energy conservation. For $\hat{H}=\frac{\hat{P}^2}{2m}+ V(\hat{X})$ we have the average
\begin{equation} \langle \psi(t)|\hat{H}|\psi(t)\rangle=E=\langle \psi(t)|\frac{\hat{P}^2}{2m}|\psi(t)\rangle+\langle \psi(t)|V(\hat{X})|\psi(t)\rangle \end{equation}
