Answer
(a)$Be+F_2\rightarrow BeF_2$
(b)$Ca+Br_2\rightarrow CaBr_2$
(c)$Ba+Cl_2\rightarrow BaCl_2$
Work Step by Step
(a)Be reacts with $F_2$ to form $BeF_2$ (Be loses two electrons and each F gains one electron)
$Be+F_2\rightarrow BeF_2$
(b)Ca reacts with $Br_2$ to form $CaBr_2$ (Ca loses two electrons and each Br gains one electron)
$Ca+Br_2\rightarrow CaBr_2$
(c)Ba reacts with $Cl_2$ to form $BaCl_2$ (Ba loses two electrons and each Cl gains one electron)
$Ba+Cl_2\rightarrow BaCl_2$
