For question (a):
To prove $P(A\cap B )= P(A)\cap P(B)$ we should show that $P(A\cap B )\subseteq P(A)\cap P(B)$ and $P(A)\cap P(B)\subseteq P(A\cap B )$.
for first part:
let take some subset $X\subseteq A\cap B$ then $X\in P(A\cap B )$. Also $X\subseteq A\wedge X\subseteq B$, means $X\in P(A)\wedge X\in P(B)$. Again this proves that $X\cap P(A)\cap P(B)$. This proves $P(A\cap B)\subseteq P(A)\cap P(B)$..................(1)
for second part:
Take any $X$ such that $X\subseteq A$ and $X\subseteq B$. This is $X\in P(A)\wedge X\in P(B)$. That means $X\in P(A)\cap P(B)$.Now also $X\subseteq (A\cap B)$. This also means that $X\in P(A\cap B)$. This proves $P(A)\cap P(B)\subseteq P(A\cap B)$ ..............(2)
From result (1) and (2),
$P(A)\cap P(B)= P(A\cap B)$
For question (b):
as obvious $sum(n)=\frac{n(n+1)}{2}$
so for n=1 $sum(1)=\frac{1(1+1)}{2}=1,$ which is true.
let assume that for n=n+m,
$sum(m+n)=sum(m)+sum(n)+mn$
$=\frac{m(m+1)}{2}+\frac{n(n+1)}{2}+mn$
$=\frac{(m+n)(m+n+1)}{2}$ is true.
then for n=m+n+1,
$sum(m+n+1)=sum(m+n)+sum(1)+(m+n)*1$
from above results,
$sum(m+n+1)=\frac{(m+n)(m+n+1)}{2}+1+(m+n)$
$=\frac{(m+n+1)(m+n+2)}{2}$
is proved by mathematical induction.