Load with a mass if 0.2kg will stretch a spring 0.1m.Spring is stretched an additional 5cm & released.Find the spring constant,period of vibration n f,max acceleration,velocity through equilibrium positions & motion equation?

Well, since the load is hanging:

#F_g = mg = -kx#

The spring constant is found for a spring at equilibrium.

Spring constant
#k = (-mg)/x = (-("0.2 kg")(-"9.80665 m/s"^2))/("0.1 m") = color(blue)("19.613 N"*"m")#

Now that you have the spring constant #k#, you can do a few more things already.

Frequency #omega# and Period #T#

#omega = sqrt(k/m)#

Well, good, we have #k# and we have #m#. So the (angular) frequency is:

#omega = sqrt("19.613kg/s"^2/"0.2 kg") = color(blue)("9.903 rad/s")#

Conveniently enough, #omega = (2pi)/T#, so the period is:

#T = (2pi" rad")/("9.903 rad/s") = color(blue)("0.634 s")#

The three equations
Next, the equations for oscillatory motion are:

#x = Acos(omegat + phi)#
#omega = -Aomegasin(omegat + phi) = A((dx)/(dt))#
#alpha = -Aomega^2cos(omegat + phi) = -omega^2x = A((d^2x)/(dt^2))#

This means the angular acceleration acts to restore equilibrium.

The max acceleration occurs at the full amplitudes since the spring is most stretched then. i.e. #x = A#. The maximum value of #cos# is #1#, so:

#color(blue)(|alpha_max| = Aomega^2)#

The max velocity occurs at the equilibrium positions since it's the middle between the two "motion boundaries" at each of which the velocity is #0#. The maximum value of #sin# is #1#, so:

#color(blue)(|v_max| = Aomega)#

And finally the motion equation is:

#color(blue)(x = Acos(omegat + phi))#

Now the only thing you really need is the amplitude. That is just the amount it was pulled, so #A = color(blue)("0.05 m")# or #"5 cm"#.