Let's check one by one
#1
And
Till now it's ok but
But as B$\cap$ C has no elements this expression yields A$\cap$ C
Hence
#2
Solve RHS
But according to first statement second part has no elements so it yields only
And
It denotes to common elements of A and B not all elements of A
Hence
#3
[tex]\\ \rm\Rrightarrow P(A-B)=32[/tex]
[tex]\\ \rm\Rrightarrow 2^{|A-B|}=32[/tex]
[tex]\\ \rm\Rrightarrow 2^{|A-B|}=2^5[/tex]
[tex]\\ \rm\Rrightarrow |A-B|=5[/tex]
[tex]\\ \rm\Rrightarrow A-B=\pm 5[/tex]
As elements can't be negative integer it's only 5.
[tex]\\ \rm\Rrightarrow A\cap B=A-(A-B)=9-5=4\star[/tex]
And
[tex]\\ \rm\Rrightarrow P(B-A)=16[/tex]
[tex]\\ \rm\Rrightarrow 2^{|B-A|}=16[/tex]
[tex]\\ \rm\Rrightarrow 2^{|B-A|}=2^4[/tex]
[tex]\\ \rm\Rrightarrow |B-A|=4[/tex]
[tex]\\ \rm\Rrightarrow B-A=\pm 4[/tex]
Again take it positive 4
[tex]\\ \rm\Rrightarrow B\cap A=B-(B-A)=7-4=3[/tex]
As
[tex]\\ \rm\Rrightarrow A\cap B\neq B\cap A[/tex]
All statements are incorrect