Seven's not an easy one to eyeball, especially on larger numbers. One solution is to mentally add the appropriate number of 7s until the number is divisible by 10, then take off the zero and repeat.
So for 5922945, you'd add 35 to get 5922980, remove the zero to get 592298.
592298 plus 42 is 592340 -> 59234
59234 plus 56 is 59290 -> 5929
5929 plus 21 is 5950 -> 595
595 plus 35 is 630 -> 63
63 plus 7 is 70 -> 7
7 is divisible by 7, so therefore 5922945 is divisible by 7.
It's slightly mentally easier if you also allow subtracting an appropriate multiple of 7 as well as adding it, because you're less likely to have to break a three-digit boundary.
So in the above case:
5922945 - 35 = 5922910
592291 - 21 = 592270
59227 - 7 = 59220
5922 +28 = 5950
595 - 35 = 560
56 is divisible by 7.
Another method is to repeatedly subtract obvious visual multiples of 7. It helps to remember that 105, 203, and 301 are multiples of 7 (differing by 98 - you can see the pattern).
+5,922,945
-5,621,742 (56, 21, 7, and 42)
=0,301,203
Which is divisible by 7.