A.
Let's see what the grammar is doing.
$S→bSe$ => For every b one e must be there
$S→PQR$ => terminals can come from either P, Q or R.
$P→bPc$ => For every b one c must be there.
$Q→cQd$ => For every c one d must be there.
$R→dRe$ => For every d one must be there.
Summary: $b-e , b-c, d-c, d-e$
So number of b and d equals number of c and e.
We can have strings $ε$, be, cd.
Hence the condition is $i+k=j+m$
where $i,j,k,m>=0$
B. We should have multiple choice for making the tree so let's try with string which can be reached using production in different ways.
String: bede
Way 1: $S→bSe$
$S→bPQRe$
$S→bPcQdRe$
$S→bcde$ (Using $P→ε$ , $Q→ε$ , $R→ε$)
Way 2: $S→PQR$
$S→bPcdRe$ (Using $Q→ε$ )
$S→bcde$ (Using $P→ε$ , $R→ε$