Is (*exptr)->cnt the same as exptr->cnt or (*exptr).cnt?

No.

(*pointer)->name says “Get the thing that pointer points to. Get the structure that it points to and get the name member from it.” For this to work, pointer must be a pointer to a pointer to a structure. For example, it could have been declared as struct foo **pointer.

pointer->name says “Get the structure that pointer points to and get the name member from it.” For this to work, pointer must be a pointer to a structure. It could have been declared as struct foo *pointer.

(*pointer).name says “Get the structure that pointer points to. Get the name member from it.” It also must be a pointer to a structure.

(The only difference between the last two is that the second uses one operator in source code. The operations actually performed are the same.)

In C, the a->b operator is a shorthand for (*a).b.

struct foo {
    int b;
};

// the . operator is used when the struct object is NOT a pointer
struct foo a;
a.b = 42;

// the -> operator is used when the struct object IS a pointer
struct foo *a = malloc(sizeof *a);
a->b = 42;

// the same thing as the above example, but using the dot operator
(*a).b = 42;

The last example is dereferencing the a pointer (getting the object it points to), then using the dot operator to access the element b inside of it. Now let's translate your question.

// first one
(*a)->b; 
// would be the same as:
(*(*a)).b;
// that is:
(**a).b;
// which would be used in
struct foo **a ... ;
(**a).b; // get the first element of the array, access field b


// second example
a->b;
// is the same as
(*a).b;
// which is your third example