Suppose for simplicity that $n = 2^N$ (otherwise, extend the array by zeroes). Maintain the following values:
$$
a_1, \ldots, a_n \\
a_1 + a_2, a_3 + a_4, \ldots, a_{n-1} + a_n \\
a_1 + \cdots + a_4, a_5 + \cdots + a_8, \ldots, a_{n-3} + \cdots + a_n \\
a_1 + \cdots + a_8, a_9 + \cdots + a_{16}, \ldots, a_{n-7} + \cdots + a_n \\
\cdots \\
a_1 + \cdots + a_{n/2}, a_{n/2+1} + \cdots + a_n \\
a_1 + \cdots + a_n
$$
By computing each line from the preceding one, you can compute all of these sums in $O(n)$.
Operation 1 can be implemented in $O(\log n)$ since each element is only involved in $\log n$ sums.
Operation 2 can be implemented in $O(\log n)$ since the sum can be broken into $O(\log n)$ sums of the type above.
Operation 3 and Operation 4 easily reduce to Operation 2 (exercise).
Details left to you.