Nodal/mesh analysis problem Consider the circuit in the attached image. The task is to determine the currents in the nodes A and B. The values of the different components are given. Any hints on where to start? I’ve considered replacing the current source with a voltage source and vice versa (the current source is connected in parallel with too many resistors to keep track of, and the voltage source isn’t connected in series with any resistor), as well as replacing some of the parallel resistors, but have been unsuccessful thus far.

 A: I assume you mean "find the voltages of nodes A and B".  Currents in node A and B doesn't make sense.
First, you can ignore $R_5$ since it is parallel with an ideal voltage source, $V_S$, and as such it will have zero affect on the rest of the circuit.
Now, you can use Bob D's suggestion and use superposition.  Kill the current source (by opening it) and then find the voltages from A to reference, and B to reference, produced by the voltage source $V_S$.  I would call these $V_Aˊ$ and $V_Bˊ$ where the single prime indicates the voltage produced by the first source we are considering, $V_S$.
Next, put the current source back, and kill the voltage source (by short circuiting it) and then find the same voltages - but this time call them $V_Aˊˊ$ and $V_Bˊˊ$.  You will see that since they are now shorted together they have the same value.
Finally, the resulting voltages with both sources in service will be,
$$V_A = V_Aˊ + V_Aˊˊ$$
and
$$V_B = V_Bˊ + V_Bˊˊ$$
Superposition can remarkably simplify many circuits problems.  To kill current sources you open them.  To kill voltage sources you short them out.
p.s.  The "reference" i'm referring to is the node at the very bottom (horizontal line).
UPDATE - showing superposition approach. The node voltages i will be finding are the voltages from the respective node to the reference (bottom of circuit).
Here is the original circuit,

First, i'll kill the voltage source (by shorting it out) and find the contribution to each nodes voltage due to the current source,

It is obvious that the currents flowing down through $R_2$ and $R_4$ have to sum to $I_S$, so we can find $V_Aˊ$ as,
$$V_Aˊ = V_Bˊ = I_S \frac{R_2R_4}{R_2+R_4}$$
Similarly, we can find $V_Cˊ$ as,
$$V_Cˊ = I_S \left(\frac{R_2R_4}{R_2+R_4} + \frac{R_1R_3}{R_1+R_3}\right)$$
Now, we kill the current source and find the contribution from the voltage source,

Using voltage division in the bottom loop (just to show a simple way),

$$V_Aˊˊ = V_S\frac{R_2}{R_2+R_4}$$
and
$$V_Bˊˊ = -V_S\frac{R_4}{R_2+R_4}$$
Now, to find $V_Cˊˊ$ i will use loop analysis which will let us find all 3 node voltages in one fell swoop.  First, i'll toss out $R_5$ since he is a useless distraction and has zero affect on the rest of the circuit,

Writing KVL loop equations we normally get simultaneous equations we can solve, but here we can solve for each loop current with just 1 equation each,
Upper loop,
$$-V_S + I_X R_1 + I_X R_3 = 0$$
So,
$$I_X = \frac{V_S}{R_1+R_3}$$
Lower loop,
$$-V_S + I_Y R_2 + I_Y R_4 = 0$$
So,
$$I_Y = \frac{V_S}{R_2+R_4}$$
And thus,
$$V_Aˊˊ = I_Y R_2 = \frac{V_S R_2}{R_2+R_4}$$
$$V_Bˊˊ = -I_Y R_4 = \frac{-V_S R_4}{R_2+R_4}$$
Then finally, we can find the voltage at node C by adding the voltage drop across $R_1$ to the voltage at node A (could have gone the other route from node B as well),
$$V_Cˊˊ = V_Aˊˊ - I_X R_1 = \frac{V_S R_2}{R_2+R_4} -  \frac{V_S R_1}{R_1+R_3}$$
$$V_Cˊˊ = V_S \left(\frac{R_2}{R_2+R_4}-\frac{R_1}{R_1+R_3}\right)$$
Our overall solution is now,
$$V_A = V_Aˊ+ V_Aˊˊ = I_S\frac{R_2R_4}{R_2+R_4}+V_S\frac{R_2}{R_2+R_4}$$
$$V_B = V_Bˊ+ V_Bˊˊ = I_S\frac{R_2R_4}{R_2+R_4}-V_S\frac{R_4}{R_2+R_4}$$
$$V_C = V_Cˊ+ V_Cˊˊ = I_S\left(\frac{R_2R_4}{R_2+R_4}+\frac{R_1R_3}{R_1+R_3}\right) + V_S \left(\frac{R_2}{R_2+R_4}-\frac{R_1}{R_1+R_3}     \right)$$
