(A) As both blocks moving together so
Time period $ = 2\pi \sqrt {\frac{{\;{\rm{m}}}}{{\;{\rm{K}}}}} ;{\rm{ where m}} = {\rm{M}} + {\rm{m}}$
$\;{\rm{T}} = 2\pi \sqrt {\frac{{{\rm{M}} + {\rm{m}}}}{{\;{\rm{K}}}}} $
(B) Let block is displaced by x in (+ve) direction so force on block will be in(-ve) direction
${\rm{F}} = - {\rm{Kx}}$
$({\rm{M}} + {\rm{m}}){\rm{a}} = - {\rm{Kx}}$
${\rm{a}} = - \frac{{{\rm{Kx}}}}{{({\rm{M}} + {\rm{m}})}}$
(C) As upper block is moving due to friction thus
${\rm{f}} = {\rm{ma}} = \frac{{{\rm{mKx}}}}{{({\rm{M}} + {\rm{m}})}}$
(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together.
$KA = (M + m)a$
$a = \frac{{KA}}{{(M + m)}}$
$f = ma = \frac{{mKA}}{{(M + m)}}$
${f_{\max }} = {f_L} = \mu {\rm{mg}}$
$f = \mu {\rm{mg}}$
$\frac{{mKA}}{{(M + m)}} = \mu {\rm{mg}}$
$A = \frac{{\mu (M + m)g}}{K}$
(E) Maximum friction can be $\mu {\rm{mg}}$ as force is acting between blocks & normal force here is mg .