Logic and Reasoning Statements

If $A\models B$ then $\mathrm{\neg }A\models \mathrm{\neg }B$

No, countermodel:
$A:=0,B:=1$ then $A\models B$ holds but $\mathrm{\neg }A\models \mathrm{\neg }B$ doesn't

If $A\models B$ and $A\wedge B\models C$ then $A\models C$

Proof by contradiction:
If this is false the first part needs to hold, while the second doesn't.
Thus $A$ and $C$ need to be $A:=1$ and $C:=0$ , because of $A⊭C$ .
For the first part to hold, as $A:=1$ , $B$ also has to be $1$ . However that means $\mathrm{\top }\models C$ , which will always be false.
$⟹$ There is no interpretation $I$ so that this statement is wrong. the statement is true

If $A\vee B\models A\wedge B$ then $A\equiv B$

Proof by contradiction:
We need an interpretation $A\not\equiv B$ i.e. $A\Rightarrow \mathrm{\neg }B$ and $B\Rightarrow \mathrm{\neg }A$ . However then $A\wedge B$ is never true, but $A\vee B$ is always true. $A\wedge B\models A\vee B$ does not hold. $⟹$ There is no interpretation $I$ so that the statement is wrong. the statement is true