Average and instantaneous velocity

The "average velocity" over a given time interval is the distance moved during that time interval divided by the length of that time interval, "change in distance"/"change in time". The "change in time" here is 0- (-2)= 2 seconds. The "change in distance" is the position at t= 0 minus the position at t= -2 (that is why lookagain asked "When x = -2, what does y equal? When x = 0, what does y equal?"). Those can be read off the graph.
 
I assumed it was part b of problem 1! Part b of problem 2 ask for the limit, as x goes to 1 from above. To answer that, ignore all parts of the graph except just above x= 1. The limit, as you approach 1 from above is -1.
 
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