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No, a Finsler metric is in general not semiriemannian. As you and JosÃ© indicate, a semiriemannian metric is always given by a positive semidefinite nondegenerate quadratic form on tangent vectors at each point in the manifold. In other words, if you fix a basis of the tangent space for a given point, then the norm of that vector is given by a homogeneous quadratic polynomial in the coefficients of the vector with respect to the basis.

On the other hand, a Finsler metric is given by a norm function on the tangent space of each point in the manifold. This norm function must be convex, and additional regularity and convexity assumptions are often made. However, there is no requirement that the norm function be given by a quadratic form. It could be given by a higher even degree polynomial in the coefficients of a vector with respect to a basis. But it could be an arbitrary sufficiently smooth sufficiently convex function, too.

One way to think about this is to consider the standard flat models. The standard flat semiriemannian model is just $R^n$ with the metric given by a non-degenerate quadratic form. The standard flat Finsler model is $R^n$ with a (sufficiently smooth and convex) Banach norm, i.e. a finite dimensional Banach space. There are obviously a lot more of the latter than the former.

One way to think about this is to consider the standard flat models. The standard flat semiriemannian model is just $R^n$ with the metric given by a non-degenerate quadratic form. The standard flat Finsler model is $R^n$ with a (sufficiently smooth and convex) Banach norm, i.e. a finite dimensional Banach space. There are obviously a lot more of the latter than the former.