Appear as complicated

Last one -- most probably 1
By congruence , 3^{φ (p)} = 1 (mod 100 ) since 3 and 100 are coprime .
Here φ (p) = 100 ( 1 - 12 ) = 50
so 3⁵⁰ = 1 (mod 100)
hence 3¹⁰⁰ = 1 (mod 100) ( squaring )
now binomially ---
( 1 + 2 )¹⁰⁰ = ¹⁰⁰ C ₀ + ¹⁰⁰ C ₁ . 2 + ....
now from second term on , all terms will be a multiple of 100 .
so last digit = ¹⁰⁰ C ₁ = 1
and second last digit = 0 ( as we get 1 + 100 . 2 +100 . something .... )
hence no. 1 is solved also

3 b > 106 = 1 ( mod 7 )
so (106)⁸⁵ = 1 (mod 7)

also 85 = 1(mod 7)
so (85)¹⁰⁶ = 1(mod 7)
so by the property of mod ,

106⁸⁵ - 85¹⁰⁶ = 0 (mod 7)
so ans --- 0

3 b . as 5 and 52 are coprime , so 5^φ(p) = 1 (mod 52)
here φ(p) = 52 ( 1- 12) ( 1 - 113 )
= 24
so 5²⁴ = 1(mod 52)
or 5⁹⁶ = 1(mod 52) (just raising to the power 4 )
or , 5⁹⁷ = 5(mod 52)

no. 2 is a very dirty sum as far as I know , so .... I will try to give the solution :)

@saumya
in post 2
φ (100) = 100 ( 1 - 1/2)(1-1/5)=40

Thanks for the answers Sawmya & eragon24

Plzzz try for 2 nd one too

another way to do Q1
3²=10-1
3¹⁰⁰=(10-1)⁵⁰=10²k-⁵⁰C₄₉10+1=100k-500+1=100(k-5)+1
So last two digits are 01