Why the polynomials with real coefficient and having degree >3 is not a vector space? In my textbook it is written that- all polynomials with real coefficient and having degree $≤3$ is a vector space. Does it also means that $>3$ are not vector space?
 A: Let $P_{n>3}$ be the set of polynomials with degree larger than 3. Then obviously, it does not contain an element that plays the role of zero vector. It is also not closed under addition. On the other hand, $P_{n\leq3}$ is a vector space because it contains the polynomial $P(X) = 0$ (zero vector), and it is closed under scalar multiplication and addition (taking a linear combination of 2 polynomials of degree 3 will never have degree larger than 3).
EDIT1: you seem to be struggling with the question:
'Why is the degree of $P(x) = 0$ zero?'
Let me try to answer that question for you.
For any constant $c \in \mathbb{R}$, we can write $P(X) = c = cX^0$. Hence, the highest power of $X$ we see here is $0$. Since $0$ is such a constant, it makes sense to say that $P(X) = 0$ has degree $0$. 
There is actually a bit ambiguity here, since we can also write, for example $P(X) = 0*X^3$, so that's why we chose to define the degree of the zero polynomial as $0$. Some authors say the degree of this polynomial is $-1$ or $- \infty$, but for these kinds of exercises it is best that you define it as being $0$ to not confuse yourself with the linear algebra concepts.
EDIT2: Now you ask why $P_{n>3}$ does not have a zero polynomial.
The easy answer is, that if it would have one, it would be $P(X) = 0$! In a vector space (or more generally a group), a subspace (or subgroup) always shares the same neutral element as the mother vector space, which here, clearly is impossible.
