How can I extract first and last line from multiple text blocks separated with new line?

One way to solve this is using a regular expression such as:

(?<testId>test\d+)(?:.*\n){4}.*(?<outcome>PASS|FAIL)

The regex matches your sample output and stores the test id (e.g. "test1") in the capture group named "testId" and the outcome (e.g. "PASS") in the capture group "outcome".

(Test it in regexr)

The regex can be used in any language with regex support. The below code shows how to do it in Python.

(Test it in repl.it)

import re

# Read from input.txt
with open('input.txt', 'r') as f:
  indata = f.read()

# Modify the regex slightly to fit Python regex syntax
pattern = '(?:.*)(?P<testId>test\d+)(?:.*\n){4}.*(?P<outcome>PASS|FAIL)'

# Get a generator which yeilds all matches
matches = re.finditer(pattern, indata)

# Combine the matches to a list of strings
outputs = ['{} {}'.format(m.group('testId'), m.group('outcome')) for m in matches]

# Join all rows to one string
output = '\n'.join(outputs)

# Write to output.txt
with open('output.txt', 'w') as f:
  f.write(output)

Running the above script on input.txt containing:

[test1]
duration
summary
code=
Results= PASS

[test2]
duration
summary=x
code=
Results=FAIL

[test444]
duration
summary=x
code=
Results= PASS

yields a file output.txt containing:

test1 PASS
test2 FAIL
test444 PASS

Using sed as tagged (although other tools would probably be more natural to use) :

sed -nE '/^\[.*\]$/h;s/^Results= ?//;t r;b;:r;H;x;s/\n/ /;p'

Explanation :

/^\[.*\]$/h         # matches the [...] lines, put them in the hold buffer
s/^Results= ?//     # matches the Results= lines, discards the useless part
t r;b               # on lines which matched, jump to label r; 
                    # otherwise jump to the end (and start processing the next line)
:r;H;x;s/\n/ /;p    # label r; append the pattern space (which contains the end of the Results= line)
                    # to the hold buffer. Switch Hold buffer and pattern space,
                    # replace the linefeed in the pattern space by a space and print it

You can try it here.

In order to print the first and last line from the block, how about:

awk -v RS="" '{
    n = split($0, a, /\n/)
    print a[1]
    print a[n]
}' input.txt

Result for the 1st example:

[Linux_MP3Enc_xffv.2_Con_37_003]
Total_Result = PASS
[Composer__vtx_007]
Total_Result = PASS
[Videox_Rotate_008]
Total_Result = PASS

The man page of awk tells:

If RS is set to the null string, then records are separated by blank lines.

You can easily split the block with blank lines with this feature.

Hope this helps.