BJT Biasing with negative voltage
Negative voltages are nothing to be afraid of. They work exactly like positive voltages, just with a minus sign. I'll walk you through the first part, which is to figure out \$R_1\$.

You know the voltage across \$R_1\$ is 15V. If you find the current through \$R_1\$, you can calculate the resistance with Ohm's Law. The current through \$R_1\$ is equal to the current through \$R_2\$ plus the base current. Ohm's Law gives you the current through \$R_2\$:

$$I_{R2} = \frac{0\ \mathrm V - -15\ \mathrm V}{5\ \mathrm{k\Omega}} = 3\ \mathrm{mA}$$

You're given the collector current and \$\beta\$, so you can easily calculate the base current:

$$I_B = \frac {I_C} {\beta} = \frac {1\ \mathrm{mA}} {100} = 0.01\ \mathrm{mA}$$

So the current through \$R_1\$ is:

$$I_{R1} = I_{R2} + I_B = 3\ \mathrm{mA} + 0.01\ \mathrm{mA} = 3.01\ \mathrm{mA}$$

This is almost (but not quite!) equal to \$I_{R2}\$ by itself. In real life, you would probably ignore \$I_B\$ unless \$\beta\$ were lower, but this is homework, so let's do it the hard way. :-)

Now you can calculate \$R_1\$:

$$R_1 = \frac {15\ \mathrm V - 0\ \mathrm V} {3.01\ \mathrm {mA}} = 4.98\ \mathrm{k\Omega}$$

I'll let you handle \$R_C\$ (trivial) and \$R_E\$ (harder) on your own. Please be sure to post the work you've done if you have any follow-up questions.