To pass from the 1st part of your exercise to the 2nd part should not be difficult.
Imagine a 3-dimentional space in which on the $x$-axis represents the $x$ projection of the velocity vector, the $y$ axis represents the y projection, and the $z$ axis the $z$ projection.
You already calculated the probability of having velocities in a small cube of dimensions $dv_x, dv_y, dv_z$, with the cube positioned at the end of the vector $(v_x, v_y, v_z)$. You got
$$q(v_x, v_y, v_z)dv_x dv_y dv_z = C exp[(−mv^2)/2kT]dv_x dv_y dv_z.$$
where $C$ is the normalization constant.
Now, you have to distinguish between probability and DENSITY of probability, i.e. probability per UNIT VOLUME of your space. The density of probability is equal to
$$C \exp[(−mv^2)/2kT].$$
Notice that this density of probability is independent of the direction of the velocity.
Now, you are asked to calculate the probability of having the velocity between $|v|$ and $|v| + d|v|$, where $|v|$ is the length of the velocity vector.
Consider therefore a sphere of radius $|v|$ in the velocity space, and around it another sphere of radius $|v| + d|v|$.
All the velocity vectors, no matter in which direction they are directed, end-up between these two spheres if their length is between $|v|$ and $|v| + d|v|$.
So, what is the volume confined between these two spheres? It is $4\pi|v|^2 d|v|$.
The density of probability (per unit of volume in the velocity space) you already calculated, and found it independent of the direction of the velocity. Then, to find the probability $dP$ in all the volume between the two spheres you multiply the probability per unit volume, with the volume. You get
$$dP = C \exp[(−mv^2)/2kT]4\pi|v|^2 d|v|.$$
What you are requested is to calculate $dP/d|v|$.
Obviously, it is equal to
$$C \exp[(−mv^2)/2kT]4\pi|v|^2.$$
Good luck,
Sofia
You have to go back to see where your equation $(1)$ came from.
For one dimension the probability of particles having a velocity between $\vec v_{\rm x}$ and $\vec v_{\rm x}+ d\vec v_{\rm x}$ is given by
$$f(\vec v_{\rm x})\;d\vec v_{\rm x} =\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,d\vec v_{\rm x}$$
The speed distribution is given by
$$f(v_{\rm x})\;dv_{\rm x} =2\,\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}=\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}$$
the factor $2$ being there because the speed (magnitude) of $\vec v_{\rm x}$ is the same as that of $-\vec v_{\rm x}$.
This is in agreement with your equation $(3)$.
Equation $(1)$ came from the idea that in three dimensions there is no preferred direction so
$f(\vec v_{\rm x},\vec v_{\rm xy},\vec v_{\rm z})\,d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z} = f(\vec v_{\rm x})d\vec v_{\rm x}\,f(\vec v_{\rm y})d\vec v_{\rm y}\,f(\vec v_{\rm z})d\vec v_{\rm z}$
You now have to count all the speeds which are the same ie the magnitude of the velocity $v$ is the same where $v^2 = v^2_{\rm x}+v^2_{\rm y}+v^2_{\rm z}$.
In this three dimensional case the volume of a shell of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z}= 4 \pi v^2 dv$ is being considered which results in your equation $(1)$.
$$f(v) \,dv = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
The equivalent distribution for two dimensions with an area of a ring of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}= 2 \pi v\, dv$ and $v^2 = v^2_{\rm x}+v^2_{\rm y}$ is
$$f(v) \,dv = \left(\frac{m}{2 \pi kT}\right)\, 2\pi v\, \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
Best Answer
The distribution is $$ w(v_x,v_y, v_z)=Ae^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}}, $$ where A is the normalization constant that can be easily found by integration (see Gaussian integral): $$ \int dv_xdv_ydv_z w(v_x,v_y, v_z) = 1. $$ The average velocity along any direction is zero, e.g., $$ \langle v_x\rangle = \int dv_xdv_ydv_z v_x w(v_x,v_y, v_z)=0 $$ This can be shown either by direct integration by parts or from the symmetry reasons: the odd function is integrated in symmetric limits (from $-\infty$ to $\infty$).
On the other hand, the speed, i.e., the magnitude of this vector, is not zero: $$ v=\sqrt{v_x^2+v_y^2+v_z^2}, \langle v\rangle =\int dv_xdv_ydv_z v w(v_x,v_y, v_z)>0, $$ since the integrand is positive everywhere. However, usually one prefers to calculate $\langle v^2\rangle$, since in this case integration is easier.
The most likely velocity corresponds (by definition) to the peak of the distribution, which here is at $v_x=v_y=v_z=0$, so it is zero as well.