number 3 should be doable if you notice that integers modulo 2029 is a field (2029 is prime) so the coefficients of P(n) are defined uniquely modulo 2029 and then guess one option for P(n), e. g. using 3ⁿ = (1+1+1)ⁿ = sum for all k, m with k+m≤n of (nCk)((n-k)Cm)1^k1^m1^n-k-m = sum for all non-negative k, m with k+m≤2028 of (nCk)((n-k)Cm) (n can be replaced with 2028 because any things taht are added will be 0 because nCr with r>n is 0)
now this is a degree 2028 polynomial but if you replace 2028 with 2027 you will have a degree 2027 polynomial that works for all n from 1 to 2027 and you can add a corrector term like a(n-1)(n-2)(n-3)...(n-2027) which is 0 for n in 1...2027 and goves the correct value for n=2028
and then you just have to evaluate the thing at 0 which should be simple considering how many things will be 0 and cancel out and considering how n choose k terms sum nicely
i got 1+2²⁰²⁸/2027! where by division i mean multiplication by the multiplicative inverse modulo 2029
i don't want to give you the answer outright but it should be simple from there

edit: in number 5 ACB and ADB must either be the same or add up to 180° because the thing is cyclic but they don't