a,
$(x - 21 \times 13) :11= 39$
$(x - 273)= 39 \times 11$
$x -273= 429$
$x= 429+ 273$
$x= 702$
b,
$(x-5) \times (1995 \times 1996 + 1996 \times 1997 )= 1234 \times 5678 \times (630- 315 \times 2) : 1996$
$(x-5) \times (1995 \times 1996+ 1996 \times 1997)= 1234 \times 5678 \times (630 - 630) : 1996$
$ (x-5) \times (1995 \times 1996 + 1996 \times 1997 )= 0$
$ x-5= 0$
$x= 5$