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.
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.
For all sets $A$ and $B, A \cup(B-A)=A \cup B$.
$$\because\quad $$ LHS $=A \cup(B-A)=A \cup\left(B \cap A^{\prime}\right)$
$\left[\because A-B=A \cap B^{\prime}\right]$
$\begin{array}{lr}=(A \cup B) \cap\left(A \cup A^{\prime}\right)=(A \cup B) \cap U & {\left[\because A \cup A^{\prime}=U\right]} \\ =A \cup B=R H S & {[\because A \cap U=A]}\end{array}$