Not an expert by any means, as I am also just beginning to learn quantum mechanics... But I will share my thoughts, and hopefully others can correct me if I am wrong.
(1) To obtain time dependent solutions for the particle in the 1-d well, we can simply tack on a time dependent function to the spatial (time independent) solution. The temporal (time) function still depends on the energy levels.
For the 1-d well, the solutions are:
$$\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
We can tack on the temporal function: $e^{\frac{-iE_n t}{\hbar}}$
But the full solution to $\Psi(x,t)$ is a superposition of states.
So $$\Psi(x,t) = \sum_{n=0}^\infty a_n \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$
And we can obtain the $a_n$ ("probability amplitudes") from the normalization condition:
$$\int_{-\infty}^\infty \mid \Psi(x,t) \mid^2 dx = 1 = \sum_{n=1}^\infty\mid a_n\mid^2$$
So in order to do this, we must be given initial conditions such as the behavior of $\Psi(x,t)$ at $t=0$. From the $\Psi(x,0)$ function we can find the $a_n$ and find how it evolves in time.
(2) Each eigenstate has quantized energy eigenvalues. The energy of the system is well defined but that does not mean the momentum is constant. Quantum mechanics is a probabilistic theory, and the momentum of a particle in a confined space is a probabilistic distribution, as is the position. You'll learn about Heisenberg's uncertainty principle, where the basic idea is that we cannot know anything about a particles momentum if it's position is completely known. In contrast, if the momentum is completely known, the position of the particle can be thought of as being spread out through all space; aka, we have infinite ignorance about the particle's position.
(3) Not entirely sure what is being asked here, but if you're asking if there is a way to determine an expectation value, then yes.
The 'distribution of the observable' can be evaluated as the expectation value of an operator, such as $x, \hat{p}, \hat{\mathcal{H}}$
Then the expectation value of $x$ will be:
$$\langle x \rangle(t) = \int_{-\infty}^\infty x~\Psi^*(x,t)\Psi(x,t)$$
This tells us the expected position (probabilistic) as a function of time, which is part of a distribution - all contained within the wave function. Essentially, the expectation value follows this probabilistic distribution.
All information about a particle is contained in its wave function $\Psi(x,t)$, but the only measurements of an operator which we can actually observe is one of its eigenvalues.
Edit: (1) There are some instances when the coefficients are all zero aside from a set number of states, such as $$\Psi(x,t) = \sum_{n=1}^2 \frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$ If $\Psi(x,0)$ was completely unknown, we cannot determine the time evolution of the wave function without first measuring a state. The time dependent Schrödinger equation is an initial value partial differential requiring us to specify a state, usually $t = 0$. But perhaps the time independent equations could be of use. It would be easy to assume the form of the wave function as a free particle: $\psi(x) = Ae^{ikx}$. Otherwise, the form of the wave function is dependent on the potential containing it (inf. sq. well, harmonic, finite well, etc).