Line 7 stores the current value that is calculated, so that the value can be used later directly without calculating it from scratch. This is memoization.

We find the nth fibonacci term by finding previous fibonacci terms, i.e. by solving subproblems. Hence, line 7 shows the optimal substructure property.

The recursive implementation doesn't store any values and calculates every value from scratch. So, the space complexity is O(1).

From the function calls, we can see that fibonacci(4) is calculated twice and fibonacci(3) is calculated thrice. Thus, the same subproblem is solved many times and hence the function calls show the overlapping subproblems property.

The recurrence relation is given by (fibo(n) = fibo(n - 1) + fibo(n - 2)). So, the time complexity is given by:

(T(n) = T(n - 1) + T(n - 2))

Approximately,

T(n) = 2 * T(n - 1)

= 4 * T(n - 2)

= 8 * T(n - 3)

:

:

:

= 2^{k} * T(n - k)

This recurrence will stop when n - k = 0

i.e. n = k

Therefore, (T(n) = 2^{n} * O(0) = 2^{n})

Hence, it takes exponential time.

It can also be proved by drawing the recursion tree and counting the number of leaves.

Consider the first five terms of the fibonacci sequence: 0,1,1,2,3. The 6th term can be found by adding the two previous terms, i.e. fibo(6) = fibo(5) + fibo(4) = 3 + 2 = 5. Therefore,the nth term of a fibonacci sequence would be given by:

fibo(n) = fibo(n-1) + fibo(n-2).

Each of the above mentioned methods can be used to find the nth fibonacci term.