Been on this problem for 2 hours now:

Four identical springs support equally the weight of a 1150kg car.

If a 91kg driver gets in, the car drops 6.8mm. What is k for each spring?

The car driver goes over a speed bump, causing a small vertical oscillation. Find the oscillation period, assuming the springs aren't damped.

The weight of the driver is supported by each spring equally so each spring experiences a force of the weight/4 which is

(91 x 9.81) / 4

= 223.1775 N

The whole car lowers by 6.8 mm and since all springs are at equal heights, each spring drops down by 6.8 mm. Using

F = kx, we get F/x = k which means k = force / extension. We need to change the units for extension which is measured in metres.

6.8 mm = 0.0068 m.

k = 223.1775/0.0068

= 32820.22 Nm^-1

This second part simple harmonic motion formulae which I'm assuming you know about.

F = ma and F = kx

ma = kx

a = kx/m since a = w^2x

w^2x = kx/m

w^2 = k/m since w = 2π/T

(2π/T) ^2 = k/m

2π/T = √ (k/m)

T = 2π x √ (m/k)

m = 1150 + 91 = 1241 kg

k = 32820.22 Nm^-1

therefore

T = 2π x √ (1241/32820.22)

= 1.22 seconds