About the nature of particle in a box Suppose we have a particle in a box, which indicates an infinitely deep potential well and we have our particle in an unequal superposition of the first two energy states, which we can write as
$$a\Psi _{1}(z,t) + b\Psi _{2}(z,t)$$
Is the probability distribution of such a system constant in time? Or is it constant in certain places in the box?
 A: 
our particle in an unequal superposition of the first two energy states, which we can write as $$a\Psi _{1}(z,t) + b\Psi _{2}(z,t)$$

The space and time dependence of the stationary states is:
$$
\Psi_n(z, t) = e^{-i E_n t/\hbar} \phi_n(z)\;,
$$
where
$$
E_n = \frac{\hbar^2 \pi^2 n^2}{2mL^2}
$$
and
$$
\phi_n(z) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi z}{L})\;,
$$
for a box from $0$ to $L$.

Is the probability distribution of such a system constant in time? Or is it constant in certain places in the box?

The probability density is:
$$
|a\Phi_1 + b\Phi_2|^2 = |a|^2|\phi_1(z)|^2 + |b|^2|\phi_2(z)|^2
+\phi_1(z)\phi_2(z)\left(ab^* e^{i\Delta t} + a^*b e^{-i\Delta t}\right)\;,\tag{1}
$$
where $\Delta = (E_2 - E_1)/\hbar$.
As seen in Eq (1) above, the first two terms are constant in time, but the last term (proportional to $\phi_1\phi_2$) changes with time.
There is one point inside the box where the probability density is constant in time. The probability density is also constant (and equal to zero) at the boundaries.
A: An eigenstate of the Hamiltonian operator is stationary (this is rather a trivial result since the operator that generates time evolution according to Schrodinger picture is basically the Hamiltonian, however make sure to look at it on a textbook). Your state is not an eigenstate of the Hamiltonian therefore (most likely) it will not be stationary i.e. it will change as time goes on (please note the words in bold, since this is not a logical consequence). The wavefunction will be zero at each of the edges since probability (the modulus squared of the wavefunction) cannot "leak out" an infinite well and must be continuous as well. You can gain some intuition and get some understanding with this simulations:
https://learncheme.com/simulations/physical-chemistry/particle-in-a-box-1/
To make the point rigorous lets write some math:
$$\Psi(t_0)=a\psi_1e^{\frac{-iE_1t_0}{\hbar}}+b\psi_2e^{\frac{-iE_2t_0}{\hbar}}.$$
Since the evolution operator is given by $U(t,t_0)=e^\frac{-iH(t-t_0)}{\hbar}$, at some later time $t$ the wavefunction will be
$$\Psi(t)=U(t,t_0)\Psi(t_0)=a\psi_1e^{\frac{-iE_1t}{\hbar}}+b\psi_2e^{\frac{-iE_2t}{\hbar}},$$
where we used the fact that: $U(t,t_0)\psi_n=\psi_ne^\frac{-iE_n(t-t_0)}{\hbar}$.
Since $E_1 \neq E_2$ a common phase factor cannot be factored out, and so clearly $\Psi(t) \neq \Psi(t_0)$, and the probability density function will pick up an oscillating cross term (the frequency of oscillation being the energy difference of the two eigenstates), so it is non constant in time.
The important point to notice is that it was not necessary to solve the Schrödinger equation; we could argue based directly on the postulates of quantum mechanics.
