Answer
(a) $^{211}_{84}Po$ $\longrightarrow$ $^{207}_{82}Pb+^{4}_{2}He$
(b) $^{207}_{81}Tl$ $\longrightarrow$ $^{207}_{82}Pb+^{0}_{-1}e$
Work Step by Step
(a) in $\alpha $ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A-4}_{Z-2}D+^{4}_{2}He$
Here $P$ is parent nucleus, $D$ is daughter nucleus and $^{0}_{-1}e$ is positron or $\beta^{-}$ particle
$^{211}_{84}Po$ $\longrightarrow$ $^{207}_{82}Pb+^{4}_{2}He$
(b)
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
$^{207}_{81}Tl$ $\longrightarrow$ $^{207}_{82}Pb+^{0}_{-1}e$
