I agree with Michael that the question is probably expecting you to estimate the answers from the graph. However, this is what I would do if these were experimental results I was analysing.
Let's assume that the blue line is a straight line $f(x)$ and the curve is a quadratic $g(x)$. A straight line has the form $y = ax + b$ where $a$ is the gradient and $b$ is the $y$ intercept. From the graph the $y$ intercept is 1 and the gradient is 4/10 i.e. 0.4. Then:
$$ f(x) = 0.4x + 1 $$
The quadratic is a little harder, but if you know the two zeroes of the quadratic, $x_1$ and $x_2$ then the function has the form $g(x) = A(x - x_1)(x - x_2)$ where $A$ is some constant. In our case the two zeros are both $x = 5$ so the function is $g(x) = A(x - 5)^2$. To find the constant $A$ note that when $x = 0$ $y = 4$ so the constant $A$ must be 4/25 or 0.16.
$$ g(x) = 0.16(x - 5)^2 = 0.16x^2 - 1.6x + 4 $$
So to find the two values of $x$ when the curves cross we just set $f(x) = g(x)$:
$$ 0.4x + 1 = 0.16x^2 - 1.6x + 4 $$
and a quick rearrangement gives:
$$ 0.16x^2 - 2x + 3 = 0 $$
To get the two solutions to this use the quadratic formula and you find the curves cross at $x \approx 1.743$ and $x \approx 10.757$.
As you say, the two particles have the same velocity when the gradients are the same i.e. $f^'(x) = g'(x)$. Differentiating our expressions for $f(x)$ and $g(x)$ gives:
$$ f^'(x) = 0.4 $$
$$ g^'(x) = 0.32x - 1.6 $$
Set these equal to find the point where the slopes are equal:
$$ 0.4 = 0.32x - 1.6 $$
$$ x \approx 6.25 $$
Incidentally, I'm a bit concerned by your calculation of the average velocity. Unless there's a bit to the question you haven't posted, the average velocity of A is distance moved (4 metres) divided by time take (10 secs) so the average velocity is 0.4 m/sec not 1.118.