How to show that a prob is in NP and that it is NP

Longest Path

We have a graph G=(V,E), lengths l(e) in Z^(+) for each e in E, a positive integer K and two nodes s,t in V.

The question is if there is a simple path in G from s to t of length at least K ?

Show that the problem Longest Path belongs to NP.
Show that the problem Longest Path is NP-complete, reducing Hamiltonian Path to it.
Show that if the graph is directed and acyclic then the problem can be solved in time O(|V|+|E|).

Could you give me a hint how we could show that the problem belongs to NP? Also, how can we reduce a problem to an other, in order to show that the latter is NP-complete?

EDIT:

So in order to show that the problem belongs to NP, do we have to draw a simple and count the sum of the lengths of the edges?

Do we say for example the following?

We see that the length of the path from the node s to the node t is equal to l((s,w))+l((w,t))=3+12=15, so there is no simple path in G from s to t of length at least K.

Or does it suffice the following?

"Given a a simple path , we can easily check if its length is at least K(by simply computing the sum of lengths of all edges in it). Thus, it is in NP."

EDIT 2: Also could you explain me further why we reduce the Hamiltonian path problem to this one in polynomial time by setting all edges' lengths equal to one and set K = |V| - 1 ?

EDIT 3: Suppose that we have a problem A and a problem B and it is known that B is NP-complete. If we want to show that A is also NP-complete, do we change the data of A in that way so that we have the same problem as the problem B and so we deduce that A is also NP-complete? Or have I understood it wrong?

EDIT 4: Also how can we show that if the graph is directed and acyclic then the problem can be solved in time O(|V|+|E|)?

EDIT 5: All edges'lengths of a Hamiltonian path are equal to 1, right? And if we have V vertices, the length of the longest path is at V-1, yes? But in our problems, the lengths of the edges aren't specific and K is also not a fixed number. So if we set all edges' lengths equal to one and set K = |V| - 1, don't we reduce our problem to the Hamiltonian path problem? Or have I understood it wrong?

To show that a problem is in NP, we need to show that it can be verified in polynomial time. Given a certificate(a simple path in this case), we can easily check that it length is at least K(by simply computing the sum of lengths of all edges in it). Thus, it is in NP.
Reduction from A to B means: given an instance of A, create an instance of B(to be more precise, we are interested in polynomial time reduction here) and solve it in order to solve the original problem. So how can we reduce the Hamiltonian path problem to this one in polynomial time? It is pretty straightforward: we can set all edges' lengths equal to one and set K = |V| - 1. Then we should try all pairs of vertices in the graph (s, t), s != t and if the solution for this problem returns true for at least one pair, return true. Otherwise, we should return false(checking that we have a path of length |V| - 1 in a graph where all edges have unit length is exactly the same thing as checking that a Hamiltonian path exists by its definition).