For all sets $A$ and $B,(A-B) \cup(A \cap B)=A$.
$\begin{aligned} \mathrm{LHS} & =(A-B) \cup(A \cap B) \\ & =[(A-B) \cup A] \cap[(A-B) \cup B] \\ & =A \cap(A \cup B)=A=\mathrm{RHS}\end{aligned}$
Hence, given statement is true.
For all sets $A, B$ and $C, A-(B-C)=(A-B)-C$.
See the Venn diagrams given below, where shaded portions are representing $A-(B-C)$ and $(A-B)-C$ respectively.
Clearly, $$ A-(B-C) \neq(A-B)-C . $$ Hence, given statement is false.
For all sets $A, B$ and $C$, if $A \subset B$, then $A \cap C \subset B \cap C$.
$$\begin{array}{ll} \text { Let } & x \in A \cap C \\ \Rightarrow & x \in A \text { and } x \in C \\ \Rightarrow & x \in B \text { and } x \in C \quad [\because A \subset B]\\ \Rightarrow & x \in(B \cap C) \Rightarrow(A \cap C) \subset(B \cap C) \end{array}$$
Hence, given statement is true.
For all sets $A, B$ and $C$, if $A \subset B$, then $A \cup C \subset B \cup C$.
$$\begin{array}{ll} \text { Let } & x \in A \cup C \\ \Rightarrow & x \in A \text { and } x \in C \\ \Rightarrow & x \in B \text { and } x \in C \quad [\because A \subset B]\\ \Rightarrow & x \in B \cup C \Rightarrow A \cup C \subset B \cup C \end{array}$$
Hence, given statement is true.
For all sets $A, B$ and $C$, if $A \subset C$ and $B \subset C$, then $A \cup B \subset C$.
$$\begin{array}{ll} \text { Let } & x \in A \cup B \\ \Rightarrow & x \in A \text { and } x \in B \\ \Rightarrow & x \in C \text { and } x \in C \quad [\because A \subset C \text { and } B \subset C]\\ \Rightarrow & x \in C \Rightarrow A \cup B \subset C \end{array}$$
Hence, given statement is true.