\(2b+1=\frac{9}{2a+1}\Rightarrow b=\frac{4-a}{2a+1}\Rightarrow b+2=\frac{4-a}{2a+1}+2=\frac{3a+6}{2a+1}\)
\(M=\frac{1}{a+2}+\frac{2a+1}{3a+6}=\frac{1}{a+2}+\frac{2a+1}{3\left(a+2\right)}=\frac{3+2a+1}{3\left(a+2\right)}=\frac{2\left(a+2\right)}{3\left(a+2\right)}=\frac{2}{3}\)