\(a.n_{H^+}=0,025.0,025.2=\dfrac{1}{800}\left(mol\right)\\ \left[H^+\right]=\dfrac{\dfrac{1}{800}}{0,1+0,025}=0,01\left(M\right)\\ \Rightarrow pH=-log\left[0,01\right]=2\\ b.n_{OH^-}=0,0025.0,1=0,00025\left(mol\right)\\ \left[OH^-\right]=\dfrac{0,00025}{0,4+0,1}=\dfrac{1}{2000}\left(M\right)\\ pH=14+log\left[\dfrac{1}{2000}\right]=10,69897\)