Answer
$a-\,\,by\,\,\,(set\,\,dif\! ference\,\,law )\\
b-\,\,by\,\,\,(set\,\,dif\! ference\,\,law )\\
c-\,\,by\,\,\,(commutative\,\,law\,\,for\,\,\cap) .\\
d-\,\,by\,\,\,(De\,morgan\,\,law )\\
e-\,\,by\,\,\,(double\,\,complement\,\,law)\\
f-\,\,by\,\,\,(distributive\,\,law)\\
g-\,\,by\,\,\,(set\,\,dif\! ference\,\,law )\\
$
Work Step by Step
$(A \cup B) - (C - A) = A \cup (B - C).\\
Proof:\,Suppose\,A,\,B,\,and\,C\,are\,any\,sets.\\ Then
(A \cup B) - (C - A) = (A \cup B) \cap (C - A)^c\\ by (set\,\,dif\! ference\,\,law) \\
= (A \cup B) \cap (C \cap A^{c})^c \\by (set\,\,dif\! ference\,\,law) \\
= (A \cup B) \cap (A^{c} \cap C)^c \\by (commutative\,\,law\,\,for\,\,\cap )\\
= (A \cup B) \cap ((A^{c})^c \cup C^{c}) \\by (De\,morgan\,\,law) \\
= (A \cup B) \cap (A \cup C^{c}) \\by (double\,\,complement\,\,law)\\
= A \cup (B \cap C^{c})\\ by (distributive\,\,law )\\
= A \cup (B - C)\\ by (set\,\,dif\! ference\,\,law) \\
$
