The [tex]n[/tex]-th term of the sequence is
[tex]a_n=8+d(n-1)[/tex]
where [tex]d[/tex] is the common difference between consecutive terms.
Since [tex]a_5=28[/tex], we have
[tex]28=8+4d\implies d=5[/tex]
so that
[tex]a_n=8+5(n-1)=5n+3[/tex]
Then the sum of the first 12 terms of the sequence is
[tex]S_{12}=\displaystyle\sum_{n=1}^{12}(5n+3)=5\sum_{n=1}^{12}n+3\sum_{n=1}^{12}1[/tex]
[tex]S_{12}=5\dfrac{12\cdot13}2+3\cdot12[/tex]
[tex]\boxed{S_{12}=426}[/tex]
