$$\begin{array}{l} \text { 1. (1) Since } \frac{a}{a+b+d}+\frac{b}{a+b+c}\frac{a}{a+b+c+d}+\frac{b}{a+b}-\frac{b}{b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1 \\ \text { Let } a \rightarrow+\infty, b=d=\sqrt{a}, c=1 \text {, the sum } \rightarrow 1, a=c \rightarrow+\infty, b=d=1 \text {, the sum } \rightarrow 2, \end{array}$$
Therefore, the range of values for $S$ is the open interval $(1,2)$.