Biasing of a circuit with two transistors BJT
We are not here to do your homework for you.

There are various ways to approach this.  Here is one.

As you say, the capacitors are open circuit for the purpose of finding the DC operating point.  That leaves two transistors and three resistors.

For the exercise, unlike in a real circuit, the gain of the transistors is explicitly known.  Once you know the voltage of the collector of Q1, everything else falls out trivially.  Start by projecting R3 back to the collector of Q1.  That becomes a load on R2.  Don't forget the B-E voltage source in series with the projected resistance.  R2 and this projected resistance together form a Thevenin source replacing R2 and the 20 V power supply.

Now think about what R1 and Q1 together look like as a load.  These can also be simplified to a Thevenin source.  Now you have two Thevenin sources connected together.  Solve for the resulting voltage.

Once you have the Q1 collector voltage, the rest follows easily.  By known that voltage, you know the current thru R1, which gives you the current thru Q1.  Likewise, you know the voltage on the emitter of Q2, which is the voltage across R3, which gives you the emitter current of Q2.

While this is a good exercise to get used to simplifying circuits one step at a time, keep in mind that in real transistor circuits the transistor gains are never that well known.  Good transistor circuits work from some minimum gain all the way to infinite gain without large changes in their operating points or characteristics.  Part of designing good transistor circuits is making sure this is true.

If I needed to find the bias point of this circuit in reality, I could not make the assumption of some nice well known gain.  I usually start with infinite gain, and see what that results in.  That's actually easier to do than any finite gain.  Then I start with the operating point found from the infinite gain approximation and change it incrementally by substituting some finite minimum gain the transistors must have.  If the result doesn't change much, then the first approximation is good enough.  If it does change a lot, then it's a crappy circuit.