How many products will be formed in this reaction?

The reaction is given as:
$(CH_3)_2C=C(CH_3)_2 + (CH_3)_3N^{+}-O^{-} + H_2O \xrightarrow{OsO_4 (10^{-4} \text{ mole})} A + (CH_3)_3N$
Product $(A)$ is:

In methyl alcohol solution,bromine reacts with ethylene (ethene) to yield $BrCH_2CH_2OCH_3$ in addition to $1, 2-dibromoethane$ because

The reaction of $4$-methyloct$-1-$ene $(P, 2.52 \ g)$ with $HBr$ in the presence of $(C_6H_5CO_2)_2O_2$ gives two isomeric bromides in a $9:1$ ratio,with a combined yield of $50 \%$. Of these,the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aqueous $K_2CO_3$ to give a non-ionic product $S$ in $100 \% $ yield. The mass (in $mg$) of $S$ obtained is. . . . . . . [Use molar mass (in $g \ mol^{-1}$) : $H=1, C=12, N=14, Br=80$]

What is the product formed in the given reaction?

Ozonolysis of the given compound yields: