circuit analysis
I have to use mesh analysis to write down the independent mesh equations. I already did this with the same circuit but with a different tree. Now I have to do it with this tree, and I have no clue how to take the meshes correctly. Can someone give me a hint?
I have tried to solve this circuit in the general case, without knowing the values for the various resistances. Just for the fun of it of course. I have applied the Extra-Element Theorem (EET, see https://en.wikipedia.org/wiki/Extra_element_theorem) with one limit though, \$V_1=V_2\$. I have used the following labels: 
The first thing is to select the extra-element, the one that bothers you or would make the analysis simpler if it were either open-circuited or replaced by a short. Here, I adopted \$R_2\$ as the extra element that I will remove (open-circuit it) from the network. I will then calculate the voltage \$V_{ab}\$ without it. This becomes my reference voltage, \$V_{ref}\$ and the final voltage applying the EET will be defined as
\$V_{ab}=V_{ref}\frac{1+\frac{R_n}{R_2}}{1+\frac{R_d}{R_2}}\$
If calculate \$V_{ref}\$ using superposition, you have
\$V_{ref}=V_1\frac{R_6}{R_6+R_5+R_4||(R_1+R_3)}(1+\frac{R_1+R_3}{R_1+R_3+R_4})\$
The second thing is to reduce the excitation voltage to 0 V, meaning you replace both sources \$V_1\$ and \$V_2\$ by a short circuit. Then, you look at the resistance offered by \$R_2\$'s terminals, again, locally applying the EET with \$R_6\$ as the extra element in this sub-circuit. 
You should find
\$R_d=(R_5+R_1||(R_3+R_4))\frac{1+\frac{R_3||((R_5||R_1)+R_4)}{R_6}}{1+\frac{R_5+R_4||(R_1+R_3)}{R_6}}\$
The last part is to find the resistance offered by \$R_2\$'s terminals when the response \$V_{ab}\$ is a null, implying that \$V_a=V_b\$. The last sketch is here 
You install a test current source \$I_T\$ which delivers across its terminals a voltage \$V_T\$. \$\frac{V_T}{I_T}\$ is the resistance you want. If you solve that circuit correctly, then you have
\$R_n=\frac{R_3(2R_1+R_5)}{2(R_1+R_3)+R_4}\$
The voltage across terminals \$a\$ and \$b\$ is finally defined as:
\$V_{ab}=V_1\frac{R_6}{R_6+R_5+R_4||(R_1+R_3)}(1+\frac{R_1+R_3}{R_1+R_3+R_4})\frac{1+\frac{\frac{R_3(2R_1+R_5)}{2(R_1+R_3)+R_4}}{R_2}}{1+\frac{(R_5+R_1||(R_3+R_4))\frac{1+\frac{R_3||((R_5||R_1)+R_4)}{R_6}}{1+\frac{R_5+R_4||(R_1+R_3)}{R_6}}}{R_2}}\$
This is a quite ugly result and it assumes that both sources are equal to form 1 single injection when nulling the response. The calculation sheet is here 
while the Mathcad using the numerical values of the original sketch gives \$V_{ab}=53.333\;V\$ and \$I=1.777\;A\$
which is the result elegantly found by jonk yesterday. I am not sure in this case the EET is the best approach, but the general expression was derived almost by inspection, except for the \$R_n\$ part which required some efforts. The EET is part of the Fast Analytical Circuits Techniques (FACTs) that allow you to derive transfer functions quickly and obtain results in a low-entropy format. You can have a look at http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202017.pdf to know more about the subject.
For a supermesh enclosing both current sources you need to come up with three equations. One equation will be a KVL equation around a loop that encloses \$I_A\$ and \$I_B\$. I see only one loop that satisfies that requirement and has no current sources on the loop itself.
The second equation states the relationship between \$I_A\$, \$I_1\$, and \$I_3\$. The third equation similarly states the relationship between \$I_B\$ and mesh currents. These last two equations you should be able to write by inspection.
Best Answer
The rules are:
So if you take R1 as link you get the mesh along R1, R2, Uq2. Then if you take R3 as link you get R3, R2, Uq2, Uq1. Finally, if you take Uq3 as link, you get your last mesh which goes around the outer perimeter.