How to eliminate the Miller effect on BJT BC547

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

1. The best way to circumvent the dependance on VA is through cascode mirroring. These have the effect of somewhat pinning the collector voltage on the output transistor so that most of the dependence on VA is removed because Vce won't be changing much on the bottom two transistors in the mirror.
2. It's an awfully low value for an early voltage, but likely not a problem for uses like a switch or a simple amplifier. The reason the early voltage is so low is because that transistor is designed for 200-300 Volts. Designing transistors to handle such extreme conditions requires design trade-offs. In this case, the gain and early voltage aren't that great.
3. Unless you're designing an integrated circuit most designers don't really have a use/need for it. You generally need matched transistors to perform what you're designing and discrete parts are hardly ever matched so adding in VA isn't going to help anything. This seems more like an issue of the right tool for the right job.
4. The impedance of the output side of a current mirror is completely dependent upon the Early Voltage as you can see here: http://users.ece.gatech.edu/phasler/Courses/ECE4430/Unit2/Current_Mirrors.pdf so there's no way around determining the output resistance if you don't know the early voltage. If you're using a simulator, you can plot the Vce vs Ic graphs and then determine the early voltage from a couple regression lines as seen here: http://en.wikipedia.org/wiki/Early_effect

In summary, I'd use different transistors to build a differential amp and go with a cascode current mirror.