analysis of two transistors ib,ic,ie?
Your are forgetting Re.

You also need to use the effective resistances for R4 and R6, which is their value 

\$R4_e = R4 * (1+\beta) = 40,000 * 101 = 4,040,000\Omega\$
\$R6_e = R6 * (1+\beta) = 1,000 * 101 = 101,000\Omega\$

Using those your base circuit now looks like this.







simulate this circuit &ndash; Schematic created using CircuitLab

You could solve that using various methods, but a shortcut is to realise that, for ideal Vbe, the voltage at the top of both emitter resistors is identical. As such you can treat both legs as being in parallel. The circuit can then be further simplified to..







simulate this circuit

Now you should be able to calculate the current through the circuit as..

\$I_{Rb} = (2.6 - 0.6)/(2,260 + 98,540) =  19.84uA\$

That makes the base voltages both..

\$V_b = 2.6 - 19.84uA * 2.26k\Omega = 2.555V\$

So 

\$Ib1 = (2.555-0.6)/4.04M\Omega = 0.484uA\$.. and
\$Ib2 = (2.555-0.6)/101k\Omega = 19.356uA\$

HOWEVER: These numbers only make sense if the transistor is not saturated.  However the question also states \$I_e \approx \beta * I_B\$ which indicates in this case they are not.