#### Answer

(a) Yes
(b) No

#### Work Step by Step

(a) $\frac{3+a}{3}=\frac{1}{3}(3+a)=1+\frac{a}{3}$ hence $\frac{3+a}{3}=1+\frac{a}{3}$
(b) When $x=2$, $\frac{2}{4+x}=\frac{2}{6}=\frac{1}{3}$ however $\frac{1}{2}+\frac{2}{x}=\frac{1}{2}+\frac{2}{2}=\frac{1}{2}+1=\frac{3}{2} $ hence$ \frac{1}{3}\ne \frac{3}{2}$ so LHS $\ne$ RHS hence $\frac{2}{4+x}\ne \frac{1}{2}+\frac{2}{x}$