a)Có $12^{2n+1}=12.144^{n}\equiv 12.11^{n} (mod133)\Rightarrow 11^{n+2}+12^{2n+1}\equiv 121.11^{n}+12.11^{n}=133.11^{n}(mod133)$ chia hết cho 133 (đpcm)b) n lẻ $\Rightarrow n$ có dạng 2k+1
$46^{n}+296.13^{n}=46^{2k+1}+296.13^{2k+1}=46.2116^{k}+296.13.169^{k}\equiv 46.169^{k}+3848.169^{k}=3894.169^{k}=1947.2.169^{k}(mod1947)$
chia hết cho 1947(đpcm)