Here is an idea. Consider the following parameterization, which is supposed to cover the configuration space in question.
$$\mathcal{C}_7:=\left\{\pmatrix{x_k\\y_x\\z_k},\pmatrix{a_k\\b_k\\c_k}_{1\leq k\leq 7}\in{\mathbb{R}^{3\times 2}}^7\,\middle |\, \text{such that conditions 1.-4. are satisfied} \right\} $$
Conditions:
- $x_k^2+y_k^2+z_k^2=1$
- $\left\langle\pmatrix{x_k\\y_k\\z_k},\pmatrix{a_k\\b_k\\c_k} \right\rangle=0$
- $a_k^2+b_k^2+c_k^2=1$
- $d(l_i,l_j)\geq 2$ for $1\leq i<j\leq 7,$ where we define the line $$l_k:=\left\{2\pmatrix{x_k\\y_k\\z_k}+\alpha\pmatrix{a_k\\b_k\\c_k}\,\middle|\,\alpha\in\mathbb{R} \right\}$$ and denote with $d(\cdot,\cdot)$ the distance between two lines.
Note that condition 4. can be rewritten as polynomial inequalities. Hence $\mathcal{C}_7$ is a semi-algebraic set in $\mathbb{R}^{42}$.
The $(x,z,y)$ are the points, where the unit cylinder is tangent to the unit sphere. The corresponding $(a,b,c)$ gives the direction in the tangent space and the lines $l$ are the cores of the cylinders. (Note that $(-a,-b,-c)$ gives the same cylinder.)
The question "Is $\mathcal{C}_7$ empty?" should be decidable. Maybe an algorithmic approach could help from here.
For the other questions the study of an analogues defined $\mathcal{C}_6$, which we know to be non-empty might be worthwhile.
I wrote a little program that tries to find points in the described semi-algebraic sets. Here's what it found for $\mathcal{C}_6$ (click here for an animation). 6 kissing cylinders http://page.mi.fu-berlin.de/moritz/mo/cyl/6.jpeg
Edit: Lets take a slightly different point of view. Fix the radius of the ball to be $1$, but let the radii of the $k$ cylinders vary while making sure all cylinders have the same radius. We can then ask: What is the largest radius $r_k$, so that we can find $k$ non-overlapping cylinders of radius $r_k$, that touch the unit ball? Hence the question is: $r_7\geq 1?$
An obvious lower bound on $r_k$ comes from the packing that allows a equatorial section which is a circle packing, as for $k=6$ in figure 1 and figure 2 in the question post. We therefore have: $$r_k\geq \frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})}$$ Here's a list of decimal approximations for small $k$s: $$\begin{array}{c|cccccc}k&3&4&5&6&7&8\\\hline \frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})} &6.464101& 2.414213& 1.425919& 1& 0.766421& 0.619914\end{array}$$ A perhaps surprising result of my calculations is the fact that $r_6>1$, indeed $$r_6> 1.04965$$ So in other words there is configuration of $6$ cylinders where the cylinders have radius larger than $1.04965$. Here is a picture of the configuration (again click here for an animation): 6 kissing cylinders http://page.mi.fu-berlin.de/moritz/mo/cyl/6b.jpeg
I also drew cylinders of radius $1$ with the same tangent points, so one can see the difference.
The configuration space can be viewed as subset of the the $6$th power of the unit tangent bundle of the sphere $(T^1(S^2))^6$ (see conditions 1.-4. and Henrik Rüping's comment).
The upshot of finding a configuration with larger radius is: the configuration space contains an open subset of $(T^1(S^2))^6$ and hence is $18$-dimensional locally.
Remark:
- For $k=5$ one can find a configuration that shows: $r_5>1.45289>1.425919$
- For $k=4$ and $k=3$ I conjecture the trivial bound for $r_k$ given above to be sharp.