1. Let the A.P. have first term $a_1=8$ and common difference $d$.

   Then the terms are

   \quad a_2=8+d, \quad a_3=8+2d, \dots, \quad a_n=8+(n-1)d.$$

2. Use the sum of the first four terms.

   $a_1+a_2+a_3+a_4=50$ $8+(8+d)+(8+2d)+(8+3d)=50$ $32+6d=50$ $6d=18$ $d=3.$

3. Now use the sum of the last four terms.

   The last four terms are: $a_{n-3},a_{n-2},a_{n-1},a_n.$

   Since $d=3$, $a_n=8+(n-1)3=3n+5.$ So, $a_{n-1}=a_n-3=3n+2,$ $a_{n-2}=a_n-6=3n-1,$ $a_{n-3}=a_n-9=3n-4.$

   Their sum is given as $170$: $(3n-4)+(3n-1)+(3n+2)+(3n+5)=170$ $12n+2=170$ $12n=168$ $n=14.$

4. Since $n=14$, the sequence has 14 terms, so the middle two terms are the 7th and 8th terms.

   $a_7=8+(7-1)3=8+18=26,$ $a_8=8+(8-1)3=8+21=29.$

5. Their product is $a_7a_8=26\times 29=754.$

Therefore, the required product is $\boxed{754}.$