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.
Imagine you have a car steering wheel connected to a heavy flywheel by a flexible, stretchy rubber band. If you turn the steering wheel slowly the flywheel starts moving and, if you are turning at a constant rate, the flywheel catches up in speed and all the energy stored in the rubber band becomes transferred to the flywheel rotation.
An inductor is the rubber band and a capacitor is the flywheel. Constant speed on the steering wheel is a dc voltage.
Now, should you rotate the steering wheel back and forth at the "right" rate, the flywheel will also start to oscillate and, in the absence of friction losses (aka resistors), that flywheel oscillation will build in amplitude and continue to build until eventually something gives out. This would be called destructive resonance and happens in electric circuits too.
The math for the electrical case and the mechanical case is virtually the same.
Should you have attached a motor instead of the steering wheel and applied a step change from zero speed to so many rpm then the flywheel would accelerate until it reaches motor rpm then continue to accelerate until all the stored energy in the rubber band was extracted. Given very low losses, the flywheel would attain a speed of double the motor speed whereupon, it starts slowing down and dumping energy back into the rubber band. At some point in time later, the flywheel will decelerate to zero speed and the process will start again.
This cycle time represents the resonant frequency of the flywheel and rubber band. In mechanical terms it relates to mass and stiffness.
Best Answer
You could consider the voltage a bit like floors on a building. A numbering system used in many places in Europe defines that the ground floor is 0 or G, that floors above it are numbered positively and numbers below it are negatively. You now have the option of measuring everything relative to ground (the floor number) or measuring the difference in level between any two floors (the potential or voltage difference).
In the left image above our man is standing on Floor 2 relative to ground. The electrical analogy is that some point on the circuit is connected to ground / earth and by convention is zero volts and all voltages (heights) are measured relative to this.
An 'all above ground' building will have no negative floors. A bunker or underground car-park will have no positive floors.
If the building is launched off into space he has no ground reference and is free to number the floors any way he wishes, including have Floor 0 at any arbitrary point. This is analogous to having an electrically isolated circuit with no ground connection in that we can call any point 'ground'.
simulate this circuit – Schematic created using CircuitLab
Figure 2. Two 1.5 V cells with three different reference points.
Hopefully Figure 2 makes it a bit clearer. Depending which point we assign as reference (GND) the other points' relative voltage changes.