Vbe₂ ≈ 656mV + 25.9mV ⋅ ln( (7.463mA) ⁄ 1mA ) ≈ 708mV
Adding that to Ve₂, I get:
Vb₂ = 7.5V + 708mV = 8.208V
I know now that the current through the collector resistor of Q₁ is:
I( 10kΩ ) = (15V - 8.208V) ⁄ (10kΩ) = 679.2μA
However, the collector current of Q₁ (Ic₁) is that value, less the base current Ib₂ required by Q₂, so:
Ic₁ = 679.2μA - 37.3μA = 641.9μA
If Q₁ is in the normal active mode, as well, then it's base and emitter currents will be:
Ib₁ = Ic₁ ⁄ β = (641.9μA) ⁄ 200 ≈ 3.21μA
Ie₁ = ((β+1)⋅Ic₁) ⁄ β ≈ 645.1μA
Now that we know Ie₁, we can compute Ve₁:
Ve₁ = Ie₁ ⋅ 1800Ω ≈ 1.161V
Again, we can estimate Vbe₁ now as:
Vbe₁ ≈ 656mV + 25.9mV ⋅ ln( (641.9μA) ⁄ 1mA ) ≈ 645mV
Adding that to Ve₁, we get the quiescent base voltage for Q₁ as:
Vb₁ = Ve₁ + Vbe₁ = 1.161V + 645mV = 1.806V
We know now the current through the 220kΩ resistor as:
I( 220kΩ ) = (15V - 1.806V) ⁄ (220kΩ) ≈ 60μA
(Hmm. Nice number.)
But we know that Ib₁ must be subtracted from that to get the current in R₁. And since we know the voltage across R₁, it's easy to now compute R₁ as:
R₁ = (1.806V) ⁄ (60μA - 3.21μA) ≈ 31.8kΩ
The above analysis answers (b) and (c).
The little-re of the 1st stage is about 40Ω, so the input resistance of the amplifier is:
R( input ) = 220kΩ || 31.8kΩ || (β+1)⋅(1.8kΩ+40Ω) ≈ 25.8kΩ
This answers (d).
The voltage gain of the emitter follower, Q₂, is only slightly less than 1:
Av₂ ≈ α ≈ β ⁄ (β+1) ≈ .995
But the voltage gain of the common emitter, Q₁, is about:
Av₁ ≈ -(10kΩ) ⁄ (1800Ω) ⋅ α
Combining these two, you get:
Av ≈ -(10kΩ) ⁄ (1800Ω) ⋅ α² ≈ -5.5
This answers (e).
If Ce were connected, Av would be quite different and would depend upon the little re value of 40Ω (because the 1800Ω value would be bypassed.)