A long double can be used if range of a double is not enough to accommodate a real number.

False, The range of float is -3.4e+38 to 3.4e+38.

True, Depending on the operating system/compiler/system architecture you are working on, the range of data types can vary.

True, When a function is declared inside the source file, that function(local function) get a priority than the extern function. So there is no need to declare a function as extern inside the same source file.

6.68 is double.
6.68L is long double constant.
6.68f is float constant.
6.68LF is not allowed in c.

int length; denotes that variable length is int(integer) data type.

char int; here int is a keyword cannot be used a variable name.

int long; here long is a keyword cannot be used a variable name.

float double; here double is a keyword cannot be used a variable name.

So, the answer is int length;(Option A).

False, The range of double is -1.7e+308 to 1.7e+308.

True,
float = 4 bytes.
double = 8 bytes.

True, we can find the size of short integer and long integer using the sizeof() operator.
Example:

#include<stdio.h>
int main()
{
    short int i = 10;
    long int j = 10;
    printf("short int is %d bytes.,\nlong int is %d bytes.",
            sizeof(i),sizeof(j));
    return 0;
}

Output:
short int is 2 bytes.
long int is 4 bytes.

Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.

Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1

Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0

Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1

Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.

Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".

Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.

Step 2: printf("The letter is"); It prints "The letter is".

Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);

The ASCII value of 'A' is 65 and 'a' is 97.

Here

=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')

=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')

=> (TRUE) ? (97): ('A')

In printf the format specifier is '%c'. Hence prints 97 as 'a'.

Step 4: printf("Now the letter is"); It prints "Now the letter is".

Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');

Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')

=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)

=> (TRUE) ? ('A') : (97)

It prints 'A'

Hence the output is

The letter is a
Now the letter is A