Why current does not flow through the 8500 ohm resistor?
Solving this circuit:

Is not an easy task because we have to solve these simultaneous equations
$$V_{CC} = (I_{E1} + I_{C2})R_C + I_{B1}R_B + V_{BE1} + V_{BE2} + I_{E2}R_{E2} $$
$$ V_{BE2} + I_{E2}R_{E2} = (I_{E1} - I_{B2})R_{E1}$$
Using this additional knowledge about &quot;internal&quot; BJT's behavior.
\$I_C = \beta I_B\$
\$I_E = I_B + I_C = I_B + \beta I_B = (\beta+1)I_B\$
\$\frac{I_C}{I_E} = \frac{\beta I_B}{ (\beta+1)I_B} = \frac{\beta}{\beta+1}\$
In fact, if we remove \$R_{E1}\$ from a circuit, this &quot;new circuit&quot; can be solved without any problem

$$V_{CC} = (I_{E1} + I_{C2})R_C + I_{B1}R_B + V_{BE1} + V_{BE2} + I_{E2}R_{E2}$$
and because now:
\$I_{E1} = (\beta_1+1)I_{B1} = I_{B2}\$
and
\$I_{E2} =(\beta_2 + 1)I_{B2}\$
\$I_{C2} =\beta_2 I_{E1}\$
We have
$$I_{B1} = \frac{V_{CC} - 2V_{BE}}{R_B + (\beta_1 + 1)(\beta_2 + 1)(R_C + R_{E2})} = \frac{3.3V - 1.4V}{2k\Omega + (101*101)(750\Omega + 1k\Omega)} = 106.4nA$$
And
\$I_{R_C} = I_{E1} + I_{C2} =  I_{E1} +  I_{E1}\beta_2 = (\beta_2 + 1)I_{E1} = (\beta_2+1)(\beta_1+1)I_{B1} = 1.086mA\$
And finally:
\$V_{OUT} = V_{CC} - I_{R_C}R_C = 2.486V\$
Now we can bring back the \$R_{E1}\$ and see how \$R_{E1}\$ influence the circuit.
We see that the \$R_{E1}\$ resistor will simply increase the \$I_{E1}\$ current.
Because now the \$I_{E1}\$ current is equal to \$I_{E1} =  I_{B2} + I_{R_{E1}}\$
And in your example :
\$I_{R_{E1}} \approx \frac{V_{B2}}{R_{E1}} \approx \frac{1.086V}{8.5k\Omega} \approx 210 \mu A\$
And this will increases the \$I_{B1}\$ to \$ 2.08 \mu A\$ and the \$Q_1\$ collector currrent to \$I_{C1} = 208 \mu A\$
Therefore
$$V_{OUT} \approx 3.3V - (210 \mu A + 1mA)750\Omega \approx 2.4V $$
As you can see not a big difference in output voltage value, so we can end-up this iterative process of finding the new operating point.