AC circuit calculations
Your answer to part three should be 
$$I=\frac{V}{Z}=\frac{1}{259.9}$$

$$P_{R_2}=I^2R_2=\frac{1}{259.9^2}50$$

The reason for this is that you have not found the complex absolute value of the impedance in your answer, (part 2 was correct, but not part 3).  You need to take the square root of the product of the complex value and it's conjugate.

$$Z=(R_1+R_2)+j \left(\frac{-1}{\omega C}\right)$$
$$\vert Z\vert=\sqrt{Z^*Z }=\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}$$

If you then use Kirchoff's Voltage Law to find the 'total voltage dropped' across the circuit, you obtain the complex expression:

$$\frac{V_0 R_1+V_0R_2+V_0\frac{-j}{\omega C}}{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}}$$

Multiplying this, by the complex conjugate and taking the square root you get

$$
\frac{
\sqrt{
V_0^2( R_1 + R_2)^2 + V_0^2\left( \frac{1}{\omega C} \right)^2 }
}{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}}
=\sqrt{V_0^2}\frac{
\sqrt{
( R_1 + R_2)^2 + \left( \frac{1}{\omega C} \right)^2 }
}{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}}=V_0$$