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 $\pm$
Best Answer
Remember: $$\cos x=1\Longleftrightarrow x=2k\pi\,,\,k\,\,\text{an integer}$$so$$\cos\left(2t+\frac{\pi}{3}\right)=1\Longleftrightarrow 2t+\frac{\pi}{3}=2k\pi$$
Assuming, as surely is the case, that it must be $\,t\geq 0\,$ , we get that$$2k\pi>\frac{\pi}{3}\Longrightarrow k=1,2,3,...$$and we can choose $k=1\Longrightarrow 2t+\frac{\pi}{3}=2\pi$