Answer
(a) $75$ ppmv
(b) $1995$
Work Step by Step
(a) To find the change in average annual $CO_{2}$ concentrations, you need to subtract the reading in 1960 from the reading in $2010$.
From the graph, we can see that:
$1960$: $312$ ppmv
$2010$: $387$ ppmv
So, now we can find the difference between the 2 readings.
Change = $387-312$
= $75$ ppmv
(b) We can look at the line for $360$ ppmv on the $y$-axis and look at what year it intersects with the blue line on the graph.
In this case, the intersection is at about halfway between $1990$ and $2000$, so the year the annual concentrations exceeded $360$ ppmv must be around $1995$.
