please solve  image  Question.

Dear Student,

There seems to be some problem in your question and appears to be incomplete as concentration at different time period is not known. 

$$\begin{gathered} \mathrm{Half} \mathrm{life} \mathrm{for} \mathrm{zero} \mathrm{order} \mathrm{reaction} \mathrm{can} \mathrm{be} \mathrm{calculated} \mathrm{as} \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{[\mathrm{A}{]}_{\mathrm{o}}}{2\mathrm{k}} \\ \mathrm{k} \mathrm{is} \mathrm{the} \mathrm{negative} \mathrm{slope} \mathrm{of} \mathrm{line} \mathrm{A} \mathrm{or} \mathrm{positive} \mathrm{slope} \mathrm{of} \mathrm{line} \mathrm{B}. \\ \mathrm{Slope}, \mathrm{k}=\frac{\mathrm{Change} \mathrm{in} \mathrm{concentration}}{\mathrm{change} \mathrm{in} \mathrm{time}} \\ \mathrm{i}.\mathrm{e}. \mathrm{if} \mathrm{concentration} \mathrm{if} \mathrm{x}, \mathrm{then} \\ \mathrm{k}=\frac{∆\mathrm{x}}{∆\mathrm{t}} \\ \mathrm{But} \mathrm{since} \mathrm{concentration} \mathrm{is} \mathrm{not} \mathrm{known}, \mathrm{so} \mathrm{k} \mathrm{cannot} \mathrm{be} \mathrm{calculated} \mathrm{and} \mathrm{hence} \mathrm{half} \mathrm{life} \mathrm{cannot} \mathrm{be} \mathrm{calculated}. \end{gathered}$$