Can someone tell me how to get the answers for 11.a) and b)

11. (a)
The instantaneous speed when t=6 sec is the slope of the curve at that point.

Did you draw the red line on the graph ?  That's absolutely the way
to do part-(a), but the line you draw should kiss the curve at the next
point ... the point on the curve where t=6, not the point where t=5.

Draw another line that just touches the curve where t=6.  Then, as well as
you can, measure the slope of that line.  You'll have to do some measuring
and some estimating.  Mark off a section of the line you draw.  Then measure
how much horizontal distance that section covers, and how much vertical distance
it covers.

The slope of the line is  (vertical distance) divided by (horizontal distance).
Its unit is  meters/second. 

11.  (b)
Finding the average speed is much easier than finding the instantaneous speed.
The average is just the slope of a line between the ends of the graph, and you
don't care how the graph twists and turns between the start and end points.

The point at the beginning is      (0 sec,  0 meters) .

The point at the end is    (10 sec,  200 meters) .

Average speed = (distance covered) divided by (time to cover the distance) .

                            =       200 meters                /                    10 seconds

                            =             20 meters/sec       from zero to 10 seconds

Depotion can i think