Q 1: $a_n= sin(\frac{n\pi}2) + \frac{(-1)^n}n, \ n \in \mathbb N$, show that , $\liminf a_n=-1$ and $\limsup a_n= 1$.
This sequence is bounded. I have found the upper and the lower bound i.e. lower bound $= -\frac 43$ and upper bound= 1 i am not able to solve it further , in an appropriate manner
I will split n into even and odd components, because n being even or odd creates different patterns for $\sin(\frac{n\pi}{2})$ and notice also if n is odd, $(-1)^n=-1$ and if n is even, $(-1)^n=1$. This means it makes a lot of sense to seperate these two ideas.
If $n$ is even, your equation becomes $$a_n=\frac{1}{n}$$ {since $\sin(k\pi)=0$ $\forall k\in\Bbb Z$}
For which the limit is $0$.
If we take n to be odd on the other hand, the scene changes quite a bit. To see the variance I will split the odd integers into two groups, those of the form $4k+1$ and those of the form $4k+3$.
This is because $\sin((\frac{4k+1}{2})\pi)=1$ for $k\in\Bbb Z$ (whole numbers) and $\sin((\frac{4k+3}{2})\pi)=-1$ for $k\in\Bbb Z$ See https://www.desmos.com/calculator/7gkcwwsibo and https://www.desmos.com/calculator/h7ubnpte1g to see this graphed, just check the value of the graph wherever $x$ is a whole number.
If $n=4k+1$, we get:
$$a_n=1-\frac{1}{n}$$ We're happy that the limit of $\frac{1}{n}$ is $0$, so this limit is $1-0=1$.
If $n=4k+3$, we get: $$a_n=-1-\frac{1}{n}$$ For which it follows that the limit is $-1$. Hence the superior(greater) limit is $1$, and the inferior (lower) limit is $-1$ as stated in your question.