BJT's two stage amplifier
Problem is I think that this first transistor is in cut off, or am I
  wrong?


If the first transistor is in cutoff, then the emitter current is zero and, thus

$$V_{E1} = -V_{DD}$$

$$V_{B1} = 0$$

So,

$$V_{BE1} = V_{DD} = 8V$$

But this is absurd (and contradicts the assumption the transistor is cutoff)!

Assuming the transistor is in forward active, KVL 'round the base-emitter loop yields:

$$V_{BE1} + I_{E1}\cdot R_{E 1}- V_{DD} + I_{B1} \cdot R_{B1} = 0$$

Can you take it from here?