2222^7777 + 7777^2222
=(2222^7)^1111 + (7777^2)^1111
The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)

2222^7 + 7777^2
=1111^2( 2^7*1111^5 + 7^2 )

If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.
We are left with 44, 77 and 99. .

11 divides 1111^2.
So the remaining term should be divisible by either 7,4 or 9
2^3 = -1(mod 9) => 2^7 = 2(mod 9)
1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2
= -2(mod 9)

2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )
9 divides ( 2^7*1111^5 + 7^2 )
Hence,99 divides the given number

Let me know if you have any problem

My take
2) 99