$X \sim U(0,1)$, so its PDF is given as: \begin{equation} f_X(x) = \begin{cases} 1 , & \text{ if } X \in ]0,1[ \\ 0 , & \text{ else } \end{cases} \end{equation}
Let $Z$ be a random variable that's independent of $X$. $Z \sim B(1,\frac{1}{2})$. Hereby we know $$P(Z=0)=P(Z=1)=\frac{1}{2}$$
How does i calculate the covariance? Cannot seem to find any rules for it?
Firstly note that $\mathbb{E}[Z] = 0 \cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{2} = \dfrac{1}{2}$. Similarly, using independence of $X$ and $Z$, we can write, $\mathbb{E}[XZ] = \mathbb{E}[X] \cdot 0 \cdot \dfrac{1}{2} + \mathbb{E}[X] \cdot 1 \cdot \dfrac{1}{2} = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}$. We also used the fact that $\mathbb{E}[X] = \dfrac{1}{2}$. And lastly, we also have that $ \mathbb{E}[XZ \cdot Z] = \mathbb{E}[XZ^2] = \mathbb{E}[X] \cdot 0 \cdot \dfrac{1}{2} + \mathbb{E}[X] \cdot 1 \cdot \dfrac{1}{2} = \dfrac{1}{4}$.
We can now use the definition of $Cov(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]$, to give that $Cov(XZ, Z) = \mathbb{E}[XZ^2] - \mathbb{E}[XZ]\mathbb{E}[Z] = \dfrac{1}{4} - \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8}$ which gives us our required value.