We have the following situation:
(60 mph-->)---------------------------------------------(60mph--->)----------------(stationary)
(*)A1<------1275.030800821355 light-years--------->A2<----------1 light-hour----->B2
---B1
---(stationary)
We are viewing the situation from the B1-B2 frame of reference.
Observers A1 and A2 are in one frame of reference.
Observers B1 and B2 are in another frame of reference.
The frame with A1 and A2 is moving 60 mph relative to the frame containing B1 and B2.
A bright flash of light is emitted, when this occurs:
A2 is 1275.030800821355 light-years from the flash.
B2 is 1275.030914898471 light-years from the flash.
A2 is 1 light-hour from B2.
A1 and B1 are right next to each other and 10 ft. from the flash.
using 186,282 mps we get 670,615,200 mph.
gamma is 1/(1-(v^2/c^2))^0.5
so gamma is 1/(1-(60/670615200)^2)^0.5 = 0.99999999999999599755180446599128.
When A2 finally arrives at B2 after 1275.030800821355 years,
A2 witnesses the flash but B2 doesn't even though A2 and B2 are at the same spot.
B2 will witness the flash an hour later.
We have a paradox involving observers A2 and B2.
Maybe we can apply time dilation and B2's clock will slow down by an hour while A2
is travelling towards B2?
1275.030800821355*0.999999999999995997551804465991 28 = 1275.0308008213498968.
so the difference is
1275.030800821355-1275.0308008213498968 = 0.0000000000051032 years.
One second is 0.00000003168808781 years.
So the difference in time between the clock at A2 and the clock at B2 when A2 reaches
B2 will be much less than one second. Time dilation doesn't resolve the paradox.
Maybe the flash in B2's frame occurred an hour earlier than the flash in A2's frame so that
they will see the flash at the same time when A2 arrives at B2?
But if we do that we now have a paradox involving observers A1 and B1. B1 will witness the flash of light but A1 won't even though they are at the same spot. A1 will see the flash an hour after B1 does. Stuck once again.
A1 and A2 are in the same reference frame and B1 and B2 are in the same reference frame.
Therefore: time@A1 = time@A2 and time@B1 = time@B2
If the flash in the B1-B2 frame occurs an hour earlier than the A1-A2 frame we get a paradox involving observers A1 and B1.
if the flash in the B1-B2 frame occurs at the same time as in the A1-A2 frame we get a paradox involving observers A2 and B2.
The flash of light represents an event and all observers within the same inertial frame will agree as to the location and time of the event.

I don't see where in any of this you have included the relativity of simultaneity. Time dilation is only part of the calculations, the Lorentz coordinate transformations include a shift in time as well as a dilation. That means two different observers can, in principle, see two spacelike events occuring in opposite orders.

We have the following situation: Your diagram is not useful.

(60 mph-->)------------------------------(60mph--->)----------------(stationary)
(*)A1<---1275.030800821355 light-years------->A2<----------1 light-hour----->B2
---B1
---(stationary)

We are viewing the situation from the B1-B2 frame of reference.
Observers A1 and A2 are in one frame of reference.
Observers B1 and B2 are in another frame of reference.
The frame with A1 and A2 is moving 60 mph relative to the frame containing B1 and B2.
A bright flash of light is emitted, when this occurs:
A2 is 1275.030800821355 light-years from the flash.
B2 is 1275.030914898471 light-years from the flash.
A2 is 1 light-hour from B2.
A1 and B1 are right next to each other and 10 ft. from the flash.
First of all, you created a mess for yourself by using a crazy mix of units.
Let's use symbols:
c = 670,615,200 mph (by your convention)
v = 60 mph
d_1 = 10 feet (irrelevant!)
d_2 = 1 light-hour = 670,615,200 miles
d_3 = (c/v)*d_2 = 7,495,412,441,184,000 miles = 11,176,920 light-hours ≈ 1275.030800821355 light-years
1 year = 365.25 * 24 * 3600 seconds (by your convention)
Events:
Flash (t_B=0, x_B = 0)
A1_flash (t_B=0, x_B= +10 feet= d_1)
B1_flash (t_B=0, x_B= +10 feet= d_1)
B2_flash (t_B=0, x_B= 7,495,413,111,799,200 miles = d_2 + d_3)
A2_flash (t_B=0, x_B= 7,495,412,441,184,000 miles = d_3)
B2_flash - A2_flash = (Δt_B=0, Δx_B = 670,615,200 miles = 1 light-hour = d_2)
(roughly consistent with 365.25 24-hour days per year)

using 186,282 mps we get 670,615,200 mph. Correct, but why are you doing such trivial multiplication?
gamma is 1/(1-(v^2/c^2))^0.5
so gamma is 1/(1-(60/670615200)^2)^0.5 = 0.99999999999999599755180446599128. Incorrect, you have calculated 1/gamma (1/\gamma)
The correct value is closer to 1.00000000000000400245

Since 0<|v|<c, it follows that 0<v²<c², 0<(v²/c²)<1, 0<1-(v²/c²)<1, 0<√(1-(v²/c²))<1, and therefore 1/√(1-(v²/c²)) > 1. If you have not internalized that gamma is greater than 1 for moving objects, then you are not experienced enough with special relativity to correctly describe physical situations.

When A2 finally arrives at B2 after 1275.030800821355 years, (Covering a distance of 1 light-hour (in B's frame) at a mutual speed of 60 mph, we see that this takes

A2meetsB2 = (t_B= 11,176,920 hours = d_2/v, x_B= 7,495,413,111,799,200 miles = d_2 + d_3)


A2 witnesses the flash but B2 doesn't even though A2 and B2 are at the same spot.
B2 will witness the flash an hour later. Completely incorrect.
Both observers think that A2atB2 happens before the light is seen by either observer.
B2seesLight = (t_B= 11,176,921 hours= (d_2 + d_3)/c, x_B= 7,495,413,111,799,200 ly = d_2 + d_3)
A2seesLight = (t_B= 11,176,921 + 1/11,176,919 hours=d_3/(c - v), x_B =
= d_2 + d_3 + v*d_2/(c - v) )

So in B's frame, the light hits B2 at a time (d_2 + d_3)/c - d_2/v = d_2/c = 1 hour later than A2meetsB2

Also in B's frame, the light hits A2 at a time d_3/(c - v) - d_2/v = d_2/(c - v) = 1.0000000894700945762 hours later than A2meetsB2

Let's talk about delta-times.
B2seesLight - A2meetsB2 = (t_B= (d_2 + d_3)/c, x_B= d_2 + d_3) - (t_B=d_2/v, x_B= d_2 + d_3) = (Δt_B= d_2/c, Δx_B=0)
A2seesLight - A2meetsB2 = (t_B= d_3/(c - v), x_B= d_2 + d_3 + v*d_2/(c - v)) - (t_B=d_2/v, x_B= d_2 + d_3) = (Δt_B=d_2/(c - v), Δx_B=v*d_2/(c - v))

What does this look like to A2.

B2seesLight - A2meetsB2 = (Δt_A = γ×d_2/c ≈ 1.000000000000004 hours, Δx_A= -γ×v×d_2/c ≈ -60.00000000000024 miles)
A2seesLight - A2meetsB2 = (Δt_A = (1/γ)×d_2/(c - v) ≈ 1.0000000894700905737 hours, Δx_A=0)
{both events happen to A2, so of course Δx has to be zero in the frame of A, but you should calculate it from the Lorentz transformations as I did... Then you should calculate the Lorentz-invariant interval and see that is the same.}

So the "paradox" is simply your mistake.