Questions about motion part 2

You haven't said on which side of 0 s is positive; I'll assume it's positive to the right of 0.
Given velocity, v(t), acceleration at time t is
a(t) = v'(t) = dv/dt

and given that and the initial position, s₀, the position at time t is

s(t) = s₀ + ∫[u=0,t] v(u) du

a) This is the acceleration at t=0, i.e., a(0)

b) Average v(t) from t₁ to t₂ is [s(t₂) - s(t₁)]/(t₂ - t₁)
This is because, to average a changing quantity from time1 to time2, you integrate it over that interval, and divide by the length of the interval.

c) Yes. This is v(5)

d) These are times, t, when v(t)=0. So you need to find the roots of that equation, and keep in mind that only those with t between 0 and 8 are valid here.

e) This will be all of the times, t, when v(t)<0

f) This will be all of the times, t, when v(t)>0

g) Speed is the magnitude of velocity, so this will be |v(4)|

h) The particle is "going faster" when it is accelerating, that is, when a(t)>0.
It is "going slower" when it is decelerating, that is, when a(t)<0.
... not necessarily when v(t)>0 or <0 [See e) and f) for those.]

i) Right. It's just
s(total) = s(8) - s(0) = ∫[u=0,8] v(u) du

a) a(t) = 2t - 9 a(0) = -9
b) x(t) = t^3/3 - 9t^2/2 + 18t + c , x(0)=1 c=1
average velocity = (x(8)-x(0))/8 = ...
c) v(5) = 5^2 - 9(5)+18 = ...