PIC16F877A+UART for LABVIEW

I need to send four measurement parameters to the PC for display and storage purposes. The software for this aim is LABVIEW. I am new to PIC and Interfacing, but by putting so much effort and asking questions from experts, i could have accomplished the acquisition of all the data till today.
I could also transmit a character "HELLO" through RS232 with my UART to USB converter and the result in hyperterminal was observed as well.
I am quite happy that i could have reached up to this stage. but on the other hand, unfortunately my knowledge is absolutely poor about LABVIEW.
I have heard that if i have control over the format of the data from PIC to LABVIEW, it is not that difficult.
Please find the attached pictures of the design for better understanding.

Questions:
1- can you please explain what is exactly meant by format of data?(I also don't know the format of data i sent through USB)
2- After specifiying the format, what would be the next stage?
3- If it is possible, please provide me with some sample codes for Hi-tech C, and LABVIEW as well.

Questions:
1- can you please explain what is the technique to transmit the 10-bit data from PIC ADC to the 8-bit serial port? something like this:

do {
for (i=0; i <=7; i++) {
myarray=read_temp();
}

for (i=0; i <=7; i++) {
putch (Lo(myarray)); //send the lower 8 bits of ADC reading
putch(Hi(myarray)); // Send upper 8 bits of ADC reading
}

2- How to identify each channel at the reception side (LABVIEW). For example, 1011011110=Channel 0 (ADC-1=Light) or 1011010110=Channel 1 (ADC-1=temperature) and so on. Something like this can work?

do {
an0 = ADC_Read(0) >> 2;
USART_Write(an0);
an1 = ADC_Read(1) >> 2;
USART_Write(an1);

an2 = ADC_Read(2) >> 2;
USART_Write(an2);

an3 = ADC_Read(3) >> 2;
USART_Write(an3);

an4 = ADC_Read(4) >> 2;
USART_Write(an4);

Thanks in advance for your help.

4 parameter, what is the maximum number for that value?
Is that 8bit is enough for the value?
ex

engshahrul said:

4 parameter, what is the maximum number for that value?
Is that 8bit is enough for the value?
ex

Code:

for(i=0;i<=3;i++) value[i]=read_a2d(i);
for(i=0;i<=3;i++) uart_transmit(value[i]);

Click to expand...

This is 10-bit value since it is produced by PIC16F877A with 10-bit resolution.

Every parameter need to divide into upperbyte and lowerbyte.

engshahrul said:

Every parameter need to divide into upperbyte and lowerbyte.

Click to expand...

i don't get the difference between uart_transmit((value>>5)&0b00011111); and uart_transmit(value&0b00011111);}
what is the meaning of 5 at here?
If you want to have the upper byte should not you &(AND) it with 0b11111000?

engshahrul said:

It shift right 5 bit.

Click to expand...

Now exactly i got your meaning. Thus, it transmit the upper byte first and lower byte second for values of i=0;i<=3. (It means 1=light, 2= temperature 3= voltage, 4= current).
Do you have any idea on how to joint this upper byte and lower byte at the other side (PC with labview)?