Answer
(a) $^{18}_{9}F$ $\longrightarrow$ $^{18}_{8}O+^{0}_{+1}e$
(b) $^{15}_{8}O$ $\longrightarrow$ $^{15}_{7}N+^{0}_{+1}e$
Work Step by Step
(a) $\beta^{+}$ deacy of $^{18}_{9}F$
in $\beta^{+}$ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A}_{Z-1}D+^{0}_{+1}e$
Here $P$ is parent nucleus,$D$ is daughter nucleus and $^{0}_{+1}e$ is positron or $\beta^{+}$ particle
$^{18}_{9}F$ $\longrightarrow$ $^{18}_{8}O+^{0}_{+1}e$
(b) $\beta^{+}$ deacy of $^{15}_{8}O$
in $\beta^{+}$ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A}_{Z-1}D+^{0}_{+1}e$
Here $P$ is parent nucleus,$D$ is daughter nucleus and $^{0}_{+1}e$ is positron or $\beta^{+}$ particle
$^{15}_{8}O$ $\longrightarrow$ $^{15}_{7}N+^{0}_{+1}e$
