\(a.\)
\(m_{NaCl}=130\cdot10\%=13\left(g\right)\)
\(m_{dd_{NaCl}}=20+130=150\left(g\right)\)
\(C\%_{NaCl}=\dfrac{20+13}{150}\cdot100\%=22\%\)
\(b.\)
\(C\%=\dfrac{S}{S+100}\cdot100\%=\dfrac{200}{200+100}\cdot100\%=66.67\%\)
\(c.\)
\(C_{M_{NaOH}}=\dfrac{0.2\cdot2+0.3\cdot1}{0.2+0.3}=1.4\left(M\right)\)