This bolded part would need to be coded in the mining software and not in the protocol, which doesn't control mining itself.
The extraNonce is a 32-bit value IIRC so a fair way to distribute all possible values equally between all the miners involves having them all go through the entire range of the other nonce, and something like 2^8 values of extraNonce as well, assuming a few million miners (the hardware) in commission and mining.
In reality the work won't be divided equally because miners aren't going to sit idly after they finish their range, because that isn't economical for them. A mining farm that pays X dollars for 5MWh of electricity a month, they're not just going to let that electricity they pay for sit wasted because it hurts their ROI.
Let's say you have a miner with double the hash rate of another miner, that faster miner will try more extraNonce values after its own range. It's this greediness around which you have to design an algorithm that completely mitigates the possibility of some other faction having greater hashpower other than the honest miners (who would also have to be defined and announced to miners via mining software).
Yeah you are right for mining software.
For the second thing you said; We solve this part with the encouragement of miners. If the miner still has not found the hash after checking its range, it will look at other ranges. In this way, they will try to find nonce in greed and race.
for the third thing you said; Nonce intervals will not be equal if the miner is too strong we will expand his nonce range. And if nonce is not within its own range it will look at other ranges as well.
With this greed there will be no untested nonce range left
It is not possible to get what you said last, we cannot do it completely, but we can increase the cost and risk of not happening. In this way, we reduce possible risks (especially for small chains). In this algorithm it does exactly that. It unites the miners as if they were a single force and also detects and prevents the attack of the attacker. The attacker will need both 51 percent to win, and plenty of luck to find two hashes.