Here, I have a problem in part b. and c. For b I found the marginal probability mass function of W and V and tried to look for a value where W<V. Should I consider P(V>W)? Are they the same thing?
In part c. I know how to calculate the mathematical expectation but here I do not understand how I should multiply W and V. I mean which particular value should I multiply??
Yes, of course $P(V>W)$ and $P(W<V)$ are the same. For b., if you consider $P(V-W>0)$ instead of $P(V>W)$, this problem will seem much simpler to you. We just need to add the probabilities, which give $V-W>0$: \begin{align} P(W<V)&=P(V-W>0)\\ &=P(V=1,W=0)+P(V=2,W=0)+P(V=2,W=1)+\cdots\\ &=22c \end{align}
For c., we shall find the conditional probability mass function, i.e., $f(z\mid V=2)=P(Z=WV=z\mid V=2).$ But this is easy thanks to the condition $\{V=2\}.$ To this end, first observe that
$$P(W=0\mid V=2)=\frac{P(W=0,V=2)}{P(V=2)}=\frac{2c}{12c}= \frac{1}{6}$$ $$P(W=1\mid V=2)=\frac{P(W=1,V=2)}{P(V=2)}=\frac{4c}{12c}= \frac{1}{3}$$ $$P(W=2\mid V=2)=\frac{P(W=2,V=2)}{P(V=2)}=\frac{6c}{12c}= \frac{1}{2}.$$ Then $$ P(Z=z\mid V=2) = \left\{ \begin{array}{ll} \frac{1}{6} & \quad z=0 \\ \frac{1}{3} & \quad z=2 \\ \frac{1}{2} & \quad z=4 \\ \end{array} \right. $$ Then we can simply compute the conditional expectation: \begin{align} E[Z\mid V=2]&=\sum_z z \ f(z\mid V=2) \\ &=\frac{1}{6}\cdot 0+\frac{1}{3} \cdot 2 + \frac{1}{2}\cdot 4\\ &=\frac{8}{3} \end{align}