Effect of a transistor on a voltage divider? (or vice versa)
The formula for the voltage at the center of a voltage divider depends on the assumption that the current through R2 is exactly equal to the current through R4.

In this case, that's not true. There's an alternate path for current to flow, through R5 and the Q1 b-e junction. 

In fact, because of it's high value, R4 is effectively irrelevant in this design. Current mostly flows through R2, R5, and the Q1 b-e.

And the results are about what you'd expect. The voltage across the Q1 b-e is about 0.7 V, and R2 and R5 are about equal.

As for the 2nd version, by reducing the current through the resistor divider that you're trying to use, you've increased the impact of "leakage" through R5 and Q1.