Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t7 + 1, y = t8 + t; t = ? 1

I must assume that you meant t=1 (not t=?1).  If t=1, here's what we'd do:

1.  Find the x and y values corresponding to t=1.  They are:
      x=(1)^7+1=2     and    y=(1)^8+1=2.  (Please note:  write t^8 instead of t8,                and write t^7 instead of t7.)
2.  The slope of the tangent line to the graph is
                          dy/dt            8t^7+1
          dy/dx =   = with 1 substituted for t
                         dx/dt              7t^6
  
      Thus, dy/dx (at t=1) = 9/7

3.  Now we have both a point (2,2) on the graph and the slope of the tangent line to the curve at that point:  9/7

4.  The tangent line to the curve at (2,2) is  found by using the point-slope formula:

y-y1 = m(x-x1)

which comes out to y-2 = 9/7(x-2), or 7y-14 = 9(x-2).  You could, if you wished, simplify this result further (e. g., by solving for y in terms of x).