I tried to sketch a simple harmonic motion below
At the top (a) the mass is moving towards the left, in this case $v<0$. Eventually it will reach the minimum value of $x$, stop, and start moving toward the right. It will get to position (b) where the location is the same as in the previous case, but the velocity has the oposite sign $v >0$. That is, for the same location $x$, the velocity could be have either positive or negative sign. As the mass keeps moving, it will get to the other side of $x = 0$ (c), in this case both $x>0$ and $v>0$. Once again it will stop at its maximum distance from the origin, change its velocity and start moving towards the left, reaching again the same position as before, but with a different sign in the velocity. I think this should answer your first question
In general for a force of the form $$F = -m\omega^2 x$$ the velocity is
$$
v^2 = \pm \left[2(E - \omega^2 x^2) \right]^{1/2}
$$
for some integration constant $E$. For another point of view, the solution to $F = m\ddot{x}$ is
$$
x(t) = A\cos(\omega t + \phi) ~ v(t) = -A\omega \sin(\omega t + \phi)
$$
where the constants $A$ and $\phi$ are integration constants. As a matter of fact $E = \omega^2 A^2/2$. Due to the oscillatory nature of the $\sin$ and $\cos$ functions, the sign of the position and velocity changes periodically.
This should answer your second question as well, if you write, for exmaple, $v = +\sqrt{(\cdots)}$, you must be certain in which part of the motion cycle. In general you should use $\pm$
You need to get the signs correct. The elastic force is always in the opposite direction to displacement. The resistance is always opposite to the velocity. Let's suppose that the equilibrium is at $x=0$. When the object is at $x=1$, the elastic force is in the negative direction. But the object might be moving towards either right or left. If it's moving in the positive direction, the resistive force is in the negative direction. If the object is moving towards the origin, the resistive force is towards the positive direction
Best Answer
Hint: Split the problem in two:
First solve the 1st order ODE $m\frac{dv}{dt}=-\beta v^n$ for the velocity $v(t)$ as a function of time $t$. (The exercise formulation doesn't say so, but there are physical reasons to believe that the velocity $v\geq 0$ should be assumed to be non-negative).
Next integrate the velocity $v(t)$ wrt. $t$ to find the position $x(t)$ as a function of time $t$.