Differential amplifier circuit, problem with saturation
I am afraid you (we) were wrong here. Just take a look at this




As you can see from the diagram we have: 

\$V_{CC} = 15V, I_{EE} = 1\textrm{mA}, Rc = 12\textrm{k}\Omega \$ 

And as you can see from the transfer characteristic the output voltage at the \$T_2\$ collector is equal to \$3V\$ for \$-1V\$ at the input (T1 base). 

\$V_{out2} = 15V - 1\textrm{mA}\cdot 12\textrm{k}\Omega = 3V\$ 

And also notice that \$T_2\$ is not in the saturation region in this case.

Because while \$T_2\$ collector is at \$3V\$. The \$T_2\$ base is at \$0V\$
Therefore, the base-collector junction is reverse biased. 

And the emitter voltage is around \$-0.7V\$ and \$V_{CE2} = 3V - (-0.7V) = 3.7V\$    

And we can easily calculate the point at which \$T_1\$ transistor enters the saturation region. For example, if we assume \$V_{CE2(sat)}=0.2V, \ V_{BE}=0.7V\$ we can solve for the minimum voltage needed at the input to saturated the \$T_1\$ transistor.   

\$U_{1(sat)} = 3V - 0.2V + 0.7V = 3.5V\$ 

And I hope that this does not need any further explanations?

But let me back to your current problem and this question:


  
  \$V_{IN1}\$ was changed to \$-1V\$. Calculate \$V_{OUT1}, \ V_{OUT2}\$
  


In this circuit, we have \$ V_{CC}=18V, Rc = 5 \textrm{k}\Omega, I_{EE} = 4\textrm{mA} \$

It should be clear for you that such a large negative voltage at the \$T_1\$ base will cut-off \$T_1\$ and allows all the \$I_{EE}\$ current to flow through the \$T_2\$ transistor (T1 cannot be a turn-on because this would cause a very large voltage across T2 base-emitter junction).

If so, the voltage at the \$T_2\$ collector is: 

\$V_{OUT2} = 18\textrm{V} - 4\textrm{mA}\cdot 5\textrm{k}\Omega = -2\textrm{V}\$ 

But notice that the \$T_2\$ base is at \$0V\$ and \$T_2\$ emitter voltage is at \$-0.7V\$. 

And the collector-emitter voltage is \$V_{CE2} = -2V - (-0.7V) = -1.3V\$  

But in NPN transistor we cannot have a negative voltage between collector and emitter (collector voltage cannot be lower than the emitter voltage).

So, this result tells us that our BJT (\$T_2\$) is in saturation region for \$-1V\$ at the input. 

Therefore: 

\$V_{out2} = V_{E2} + V_{CE(sat)} = -0.7+0.2V = -0.5V\$ 

And the \$T_2\$ collector current is 

\$ I_{C2(sat)} = \frac{V_{CC} - V_{out2} }{R_{C2}} = 3.5\textrm{mA}\$ 

The base current is \$I_{B2} = I_{EE} -I_{C2(sat)} = 4\textrm{mA} - 3.5\textrm{mA} = 0.5\textrm{mA} \$

I hope that this explains why \$T_2\$ is indeed in the saturation region for \$-1V\$ at the input (T1 base). Because here we have a different situation than shown in your example transfer characteristic.