another acceleration with velocity questions

Please limit to two questions per thread. I have deleted #3 and #4.

1) The dog acceleration is a. The dog accelerates until he reaches the speed of 20m/s (having covered x=35m). After that, he maintains the speed of 20m/s untill he reaches the =100m.

Firstly we find the acceleration:
we have: u=a*t1 <=> 20=a*t1 <=> t1= 20/a
and
x=(1/2)a*t12 <=> 35=(1/2)*a(20/a)2 <=> 35=(1/2)*a*(400/a2) <=> 35=(1/2)*400/a <=> 35=200/a <=> a=200/35 <=> a=5,71m/s2
and t1=20/a=3,5s

This acceleration applies until the dog reaches x=35 (t=3,5s)
(note: we assume that the acceleration remains constant until he reaches its max speed)

Then, we need to find how much time the dog need to reach the 100m
The dog is located at x=35 and must cover the distance of 100-35=65m. It's speed is constant: u=20m/s. This is simpler: x=u*t <=> t=x/u=65/20=3.25s (at last, a number that doesn't need rounding)
So, t2=t1+t= 3,5+3,25=6,75s. (t2 is the total time when the dog reaches 100m)

I'll probably check the rest of your questions tomorrow.