Is the inverse of a symmetric matrix also symmetric?

You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more complete proof. Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $:

$$ I = I^T $$

since $ AA^{-1} = I $,

$$ AA^{-1} = (AA^{-1})^T $$

since $ (AB)^T = B^TA^T $,

$$ AA^{-1} = (A^{-1})^TA^T $$

since $ AA^{-1} = A^{-1}A = I $, we rearrange the left side

$$ A^{-1}A = (A^{-1})^TA^T $$

since $ A = A^T $, we substitute the right side

$$ A^{-1}A = (A^{-1})^TA $$ $$ A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$ A^{-1}I = (A^{-1})^TI $$ $$ A^{-1} = (A^{-1})^T $$

and we are done.

In fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.

Yes.

$$ AB=BA=I\quad\Rightarrow\quad B^TA^T=A^TB^T=I\quad\Rightarrow\quad B^TA=AB^T=I $$