Let us assume the following probabilities:

|$Y$|$X=1$|2|3||$1$|$p_1$|$p_2$|$p_3$||$2$|$p_4$|$0$|$p_5$||$3$|$0$|$p_6$|$0$|

From the fact that the sum of all entries is $1$, we have that$$p_1+p_2+p_3+p_4+p_5+p_6=1.$$

Using (a)
for $k=1$, $$\frac{p_1}{p_1+p_4}=\frac{2}{3}\implies p_1=2p_4$$
for $k=2$, $$\frac{p_2}{p_2+p_6}=\frac{2}{3}\implies p_2=2p_6$$
for $k=3$, $$\frac{p_3}{p_3+p_5}=\frac{2}{3}\implies p_3=2p_5$$

So,$$p_1+p_2+p_3=2(p_4+p_5+p_6)=2(1-p_1-p_2-p_3)\implies p_1+p_2+p_3=\frac{2}{3}.$$

Using (b)
for $k=1$, $$\frac{p_1}{p_1+p_2+p_3}=\frac{1}{6}\implies p_1=\frac{1}{6}(p_1+p_2+p_3)=\frac{1}{9}$$
for $k=2$, $$\frac{p_2}{p_1+p_2+p_3}=\frac{2}{6}\implies p_2=\frac{2}{6}(p_1+p_2+p_3)=\frac{2}{9}$$
for $k=3$, $$\frac{p_3}{p_1+p_2+p_3}=\frac{3}{6}\implies p_3=\frac{3}{6}(p_1+p_2+p_3)=\frac{3}{9}$$

Also, $$p_4=\frac{1}{18},p_5=\frac{3}{18},p_6=\frac{2}{18}.$$