$27^{27^{26}}=(3^3)^{3^{3.26}}=(3^3)^{3^{78}}=3^{3^{79}}$
To find the last digit, we need to take modulo 10.
By Euler's totient theorem, $3^{\phi(10)}≡1(mod\ 10)=>3^4≡1(mod\ 10)$
Now $3^{79}=(4-1)^{78} 3≡3(mod\ 4)=4c+3$ for some integer c.
So, $27^{27^{26}}=3^{3^{79}}=3^{4c+3}=(3^4)^c\cdot 3^3≡1\cdot 27(mod\ 10)≡7(mod\ 10)$
To find the last two digit, we need to take modulo 100.
(1)
(i)Now by Euler's totient theorem, $\phi(4)=2=>3^2≡1(mod\ 4)$ and $\phi(25)=20=>3^{20}≡1(mod\ 25)$
$=>3^{lcm(2,20)}≡1(mod\ 100)=>3^{20}≡1(mod\ 100)$
So, we need find $3^{79}(modulo\ 20)=d$(say)
As $\phi(20)=\phi(5)\phi(4)=4.\cdot 2=8$
So, $3^{8}≡1(modulo\ 20)=>3^{80}≡1=>3(3^{79})≡1=>3d≡1(modulo\ 20)$ as $d=3^{79}(modulo\ 20)$
$=>d≡7(modulo\ 20)=20e+7$(say) for some integer e=>$3^{79}=20e+7$.
$=>27^{27^{26}}=3^{3^{79}}=3^{20e+7}=(3^{20})^e\cdot 3^{7}≡3^7(mod\ 100)=3^6\cdot 3≡29\cdot 3$ (As $3^6=729$)
$≡87(mod\ 100)$
(ii) Alternatively using Carmichael Function , we find that $3^{20}≡1(mod\ 100)$ as $λ(100)=(λ(25),λ(4))=(20,2)=20$.
Rest will be same as (i)
........
(2)
As 100=25*4 where (25, 4)=1 , let us find $3^{3^{79}}(modulo\ 4)$ and $3^{3^{79}}(modulo\ 25)$
As $3^{79}$ is odd , $3^{3^{79}}≡(4-1)^{3^{79}}≡3(mod\ 4)$
As $\phi(25)=20$, we need to find $3^{79}(modulo\ 20)$ which we have already found to be of the form 20e+7.
$3^{3^{79}}(modulo\ 25)=3^{20e+7}=(3^{20})^e\cdot 3^7 ≡3^7(mod\ 25)=(27)^2\cdot 3≡2^2\cdot 3(mod\ 25)≡12(mod\ 25)$
So, we need to find x such that $x≡12(mod\ 25)$ and $x≡3(mod\ 4)$
So, $x=25p+12=4q+3$ for some integers p,q.
Or, $25p=4q-9=4q-(25-6.4)9$
$=>25(p+9)=4(q+54)$
$=>4|(p+9)=>4|(p-3)=>p=4r+3$ for some integer r.
So,$x=25p+12=25(4r+3)+12=100r+87$