Answering questions about the recurrence of certain aspects of an algorithm
Firstly, there is one minor typo in the question. The "$T(n) =\frac 32 * T(n-1) + 1$" should be "$T(n) =\frac 32 * (T(n-1) + 1)$" since this equality is said to be obtained by simple algebra from the equality above it. However, this typo does not affect either the question or this answer.



The answer to all of your questions are so simple that I can hardly imagine that you missed it. Well, on the other hand, it is indeed very easy to miss it. 

Let me give you a simple example to illustrate the answer. Suppose we have the following function, which computes the factorial.

function FACTORIAL(n)
    if n = 1 then
        return 1
    else
        return  n * FACTORIAL(n-1)
    end-if


Suppose we have made a call, FACTORIAL(4). That means we must have called FACTORIAL(3), FACTORIAL(2) and finally FACTORIAL(1). So we have made four calls to this function in total, the last three of which are called recursively. In fact, we have the following recurrence relations for $U(n)$, the number of recursive calls made in FACTORIAL(n).
$$U(1) = 0$$
$$U(n) = 1 + U(n-1), \text { for } n\gt 1$$
Notice the first "1" in the last equality, which stands for the call to FACTORIAL with parameter $n-1$ in the code, n * FACTORIAL(n-1) while $U(n-1)$ comes from the number of recursive calls made by FACTORIAL(n-1). In other words, the recursive calls made by FACTORIAL(n-1) does not include that call itself, which is, though of course, counted towards the recursive calls made by FACTORIAL(n). We can also check by contradiction. Let us suppose we did not have that "1". Then we would get $U(4)=U(3)=U(2)=U(1)=0$, which said that we had made no recursive calls to FACTORIAL when we had computed FACTORIAL(4), which is not true at all.

In the same way as in this simple example, all those constant 1's and one 2 in the question come from right where the calls are made. For example, when x = 2, the function will return (3*RANDOM(n-1) + RANDOM(n-1) + 1), which incurs $(1+R(n-1)) + (1+ R(n-1)) = 2 + 2*R(n-1)$ recursive calls.