Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\arccos\left(\frac{6\sqrt{2}-4}{15}\right)$$ and
$$x_2=\arcsin\left(\frac{3+8\sqrt{2}}{15}\right)$$
but how to compare now?

(I'm going to do this in degrees rather than radians as this is very intuitive about right triangles).
Obviously if $\theta = \arccos \frac ah$ then $\theta$ represents an angle of a right triangle with adjacent side $a$ and hypotenuse $h$ (and opposite side $o = \sqrt {h^2 - a^2}$).  So $\arcsin \frac ah = 90 - \theta$, the other angle of the same triangle where the side $a$ is now opposite rather than adjacent side.
[These are the triangles with sides $3,4,5$ and with sides $1, 2\sqrt{2}, 3$.]
So if $x <:> \frac{\sqrt{2}}{2}$ then $\arccos x = 45 \pm \phi$ for some positive angle $\phi$ and $\arcsin x = 45 \mp \phi$.  (Draw a picture.  It is obvious.)
So $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta$
And $\arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
So to solve this problem is simple and matter of figuring out which is larger $\phi$ or $\theta$.  I.E. which triangle has the steeper slope the the $3$, $4$, $5$ triangle, or the $1, 2\sqrt{2}, 3$ triangle.  Then answer is obviously the $1, 2\sqrt{2}, 3$ triangle 
and so $\theta > \phi$ and $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta > \arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
===== or even more straightforward =====
$\sin x = \sqrt {1 - \cos^2 x} =y$
$\arcsin y = \arccos (\sqrt{1 - y^2})$
So $x_1 = \arccos \frac 35 + \arccos \frac{2\sqrt2}{3}= \arcsin \sqrt{ 1- \frac 35^2} + \arcsin \sqrt{1- \frac {2\sqrt{2}}3^2} = \arcsin \frac 45 + \arcsin \frac 13$
$x_2 - x_1 = (\arcsin \frac  {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45)$
Now as $0 < 1/3 < 3/5  < 4/5 < \frac {2\sqrt 2} 2 < 1$ so $0 < \arcsin 1/3 <\arcsin 3/5  < \arcsin4/5 <\arcsin \frac {2\sqrt 2} 2 < 1$
So $x_2 - x_1 = (\arcsin \frac  {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45) > 0$.
So $x_2 > x_1$.
