Đáp án:

$\begin{array}{l}
a)m = 2\\
 \Rightarrow \left\{ \begin{array}{l}
3x - y = 3\\
x + y = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
4x = 3 + 2\\
x + y = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{4}\\
y = 2 - x = 2 - \dfrac{5}{4} = \dfrac{3}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{4};y = \dfrac{3}{4}\\
b)\left\{ \begin{array}{l}
\left( {m + 1} \right).x - y = m + 1\\
x + \left( {m - 1} \right).y = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left( {m + 1} \right)x - y = m + 1\\
\left( {m + 1} \right).x + \left( {m - 1} \right)\left( {m + 1} \right).y = 2\left( {m + 1} \right)
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left( {m + 1} \right).x - y = m + 1\\
\left( {m + 1} \right).x + \left( {{m^2} - 1} \right).y = 2m + 2
\end{array} \right.\\
 \Rightarrow \left( {{m^2} - 1} \right).y + y = 2m + 2 - m - 1\\
 \Rightarrow {m^2}.y = m + 1\\
 + Khi:m = 0 \Rightarrow 0.y = 1\left( {ktm} \right)\\
 \Rightarrow \text{phương trình vô nghiệm}\\
 + Khi:m \ne 0 \Rightarrow y = \dfrac{{m + 1}}{{{m^2}}}\\
 \Rightarrow x = 2 - \left( {m - 1} \right).y\\
 = 2 - \left( {m - 1} \right).\dfrac{{m + 1}}{{{m^2}}}\\
 = 2 - \dfrac{{{m^2} - 1}}{{{m^2}}} = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
c)m \ne 0\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}} = 1 + \dfrac{1}{{{m^2}}}\\
y = \dfrac{{m + 1}}{{{m^2}}}
\end{array} \right.\\
x \in Z;y \in Z\\
 \Rightarrow \dfrac{1}{{{m^2}}} \in Z\\
 \Rightarrow {m^2} = 1\\
 \Rightarrow \left[ \begin{array}{l}
m = 1 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2\left( {tm} \right)
\end{array} \right.\\
m =  - 1 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 0\left( {tm} \right)
\end{array} \right.
\end{array} \right.\\
Vậy\,m = 1;m =  - 1\\
d)x + y\\
 = 1 + \dfrac{1}{{{m^2}}} + \dfrac{{m + 1}}{{{m^2}}}\\
 = 1 + \dfrac{1}{{{m^2}}} + \dfrac{1}{m} + \dfrac{1}{{{m^2}}}\\
 = 2.\left( {\dfrac{1}{{{m^2}}} + \dfrac{1}{2}.\dfrac{1}{m} + \dfrac{1}{{16}}} \right) + 1 - \dfrac{1}{8}\\
 = 2.{\left( {\dfrac{1}{m} + \dfrac{1}{4}} \right)^2} + \dfrac{7}{8} \ge \dfrac{7}{8}\\
 \Rightarrow \dfrac{1}{m} =  - \dfrac{1}{4}\\
 \Rightarrow m =  - 4\left( {tm} \right)
\end{array}$