Replace each tab ONLY at the beginning of each line with spaces

Sed matches entire lines but doesn't include the newline, so a blank line will just be an empty string. You can use ^ to match the beginning of a line and $ to match the end, so ^$ matches a blank line. Just replace that with % ghi\n%:

sed 's/^$/% ghi\n%/'

The newline that already existed will remain, so you'll end up with % ghi on one line and % on the next

Edit: If it needs to only match once then the expression is a bit more complicated. The easiest way I know to do it in sed is:

sed '0,/^$/ s/^$/% ghi\n%/'

The replacement is wrapped in an address range 0,/^$/, which means "only apply the following to the lines between 0 and the first line that matches ^$". Since the replacement expression checks for ^$ as well, the only line that's actually going to change is the first that matches ^$ -- the first blank line

Using gsub() in awk on the last ;-delimited field:

$ awk -F ';' 'BEGIN { OFS=FS } { gsub(" ", ",", $NF); print }' file
some name;another thing; random;,value,value,value,value,value

Using sed and assuming we'd like to replace all spaces after the last ; with commas:

$ sed 'h;s/.*;//;y/ /,/;x;s/;[^;]*$//;G;s/\n/;/' file
some name;another thing; random;,value,value,value,value,value

Annotated sed script:

h               ; # Duplicate line into hold space
s/.*;//         ; # Delete up to the last ;
y/ /,/          ; # Change space to comma in remaining data
x               ; # Swap pattern and hold spaces
s/;[^;]*$//     ; # Delete from the last ;
G               ; # Append hold space delimited by newline
s/\n/;/         ; # Replace the embedded newline with ;
                ; # (implicit print)

The "hold space" is a separate storage buffer that sed provides. The "pattern space" is the buffer into which the data is read from the input and to which modifications may be applied.