If $$a \colon (b+c) = 1 \colon 3$$ and $$c \colon (a+b) = 5 \colon 7$$, then $$b \colon (c+a)$$ is

Let $$S=a+b+c$$. From the first ratio, $$b+c=3a$$, so $$S=4a$$. From the second, $$7c=5(a+b)=5(S-c)$$, so $$12c=5S$$, giving $$c=\frac{5S}{12}$$. Then $$b=S-a-c=S-\frac{S}{4}-\frac{5S}{12}=\frac{S}{3}$$, and $$c+a=\frac{5S}{12}+\frac{S}{4}=\frac{2S}{3}$$. So $$b \colon (c+a) = \frac{S}{3} \colon \frac{2S}{3} = 1 \colon 2 = 0.5$$.