(a)
Sentential form $A_{n}$ of length $n$ has exactly $1$ non-terminal (i.e $S$), and has form $a^{i}Sb^{j} (0 \leq i,j)$.
From Production rules of grammar it's easy to see that after substituting any of first two productions,
form of Sentential form $A_{n+1}$ remains $a^{i+1}Sb^{j} (0 \leq i,j)$ or $a^{i}Sb^{j+1} (0 \leq i,j)$.
Upon applying production 3 or 4 form of string derived will have form $a^{i+1}b^{j} (0 \leq i,j)$ or $a^{i}b^{j+1} (0 \leq i,j)$ respectively.
For sentential form of length 1 our assumption that it has form $a^{i}Sb^{j} (0 \leq i,j)$ is true.
Now because every string derived from grammar have form $a^{i+1}b^{j} (0 \leq i,j)$ or $a^{i}b^{j+1} (0 \leq i,j)$, we can say that substring $ba$ will never occur in any string derived from this grammar.
(b)
language generated by given grammar is set of all strings having arbitrary number of a's followed by arbitrary number of b's.