The DC collector current is determined by \$R_E\$:
\$I_C = \alpha \dfrac{9.4V}{R_E} \approx \dfrac{9.4V}{R_E}\$
Since you require \$I_C < 1.25mA \$, the constraint equation is:
\$R_E > \dfrac{9.4V}{1.25mA} = 7.52k\Omega\$
The second requirement, maximum output voltage swing, without any other constraint, doesn't fix the collector resistor value.
We have:
\$ V_{o_{max}} = 19.8V - I_C(R_C + R_E)\$
But, the voltage across \$R_E\$ is fixed at 9.4V so:
\$V_{o_{max}} = 10.4V - I_C R_C\$
\$V_{o_{min}} = -I_C * R_C||R_L\$
If you stare at this a bit, you'll see that maximum output voltage swing is 10.4V but this requires that the product \$I_C R_C = 0\$* which is absurd.
Now, if we also require symmetric clipping, then, by inspection:
(1) \$V_{o_{max}} - V_{o_{min}} = 2 I_C (R_C||R_L)\$
(2) \$10.4V = I_C(R_C + R_C||R_L) \$
Looking at (1), note that, for maximum swing, we get more "bang for the buck" by increasing \$I_C \$ rather than \$R_C \$.
Since we have an upper limit on \$I_C\$, (2) becomes:
\$R_C + R_C||R_L = \dfrac{10.4V}{1.25mA} = 8.32k \Omega\$
which can be solved for \$R_C\$.
*unless \$R_L\$ is an open circuit
I don't now if your simulation is correct, but you could make it less beta-dependant by increasing negative local feedback, by for example emitter degradation, such as increasing R1 and R2.
By using a Darlington or Sziklai pair, R4 and R5 can be increased while still maintaining output current. Discussion of thermal properties: https://sound-au.com/articles/cmpd-vs-darl.htm#s4
If some crossover distortion is acceptable, the biasing voltage could be lowered so that the quiescent current through Q1, Q2 is 0.
Best Answer
Negative voltages are nothing to be afraid of. They work exactly like positive voltages, just with a minus sign. I'll walk you through the first part, which is to figure out \$R_1\$.
You know the voltage across \$R_1\$ is 15V. If you find the current through \$R_1\$, you can calculate the resistance with Ohm's Law. The current through \$R_1\$ is equal to the current through \$R_2\$ plus the base current. Ohm's Law gives you the current through \$R_2\$:
$$I_{R2} = \frac{0\ \mathrm V - -15\ \mathrm V}{5\ \mathrm{k\Omega}} = 3\ \mathrm{mA}$$
You're given the collector current and \$\beta\$, so you can easily calculate the base current:
$$I_B = \frac {I_C} {\beta} = \frac {1\ \mathrm{mA}} {100} = 0.01\ \mathrm{mA}$$
So the current through \$R_1\$ is:
$$I_{R1} = I_{R2} + I_B = 3\ \mathrm{mA} + 0.01\ \mathrm{mA} = 3.01\ \mathrm{mA}$$
This is almost (but not quite!) equal to \$I_{R2}\$ by itself. In real life, you would probably ignore \$I_B\$ unless \$\beta\$ were lower, but this is homework, so let's do it the hard way. :-)
Now you can calculate \$R_1\$:
$$R_1 = \frac {15\ \mathrm V - 0\ \mathrm V} {3.01\ \mathrm {mA}} = 4.98\ \mathrm{k\Omega}$$
I'll let you handle \$R_C\$ (trivial) and \$R_E\$ (harder) on your own. Please be sure to post the work you've done if you have any follow-up questions.