jmc

By Vieta's formulas, $a+b+c=7,$ $ab+ac+bc=5,$ and $abc=-2.$

We can say $$\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}=\frac{a^2}{abc+a}+\frac{b^2}{abc+b}+\frac{c^2}{abc+c}.$$Since $abc=-2,$ this becomes $$\frac{a^2}{a-2}+\frac{b^2}{b-2}+\frac{c^2}{c-2}.$$By Long Division, $\frac{x^2}{x-2}=x+2+\frac{4}{x-2},$ so $$\begin{array}{rl}\frac{a^2}{a-2}+\frac{b^2}{b-2}+\frac{c^2}{c-2}& =a+2+\frac{4}{a-2}+b+2+\frac{4}{b-2}+c+2+\frac{4}{c-2}\\ & =a+b+c+6+4\left(\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2}\right)\\ & =7+6+4\cdot \frac{(b-2)(c-2)+(a-2)(c-2)+(a-2)(b-2)}{(a-2)(b-2)(c-2)}\\ & =13+4\cdot \frac{(ab+ac+bc)-4(a+b+c)+12}{abc-2(ab+ac+bc)+4(a+b+c)-8}\\ & =13+4\cdot \frac{5-4\cdot 7+12}{-2-2\cdot 5+4\cdot 7-8}\\ & =\boxed{\frac{15}{2}}.\end{array}$$