(1)
The second rank tensor given by \(a_ib_j dx^i \otimes dx^j\) has the value on vectors \(v\) and \(w\):
$$a_ib_j dx^i(v) \otimes dx^j(w) = a_iv^i b_jw^j.$$
For one-forms \(\alpha\) and \(\beta\) with components \(a_i\) and \(b_j\), the tensor product \(\alpha \otimes \beta\) has the value on vectors \(v\) and \(w\):
$$\alpha(v) \otimes \beta(w) = a_iv^i b_jw^j.$$
Thus,
$$\alpha \otimes \beta = a_ib_j dx^i \otimes dx^j.$$
(2)
let \(\mathbf{A} : E \to E\) be a linear transformation.
(i) The trace tr(\(\mathbf{A}) = A_i^i\). If we change coordinate systems from \(x\) to \(y\) the components of the mixed tensor of \(\mathbf{A}\) transform as
$$A'^l_k = \frac{\partial x^j}{\partial y^k}\frac{\partial y^l}{\partial x^i}A^i_j.$$
Therefore, tr(\(\mathbf{A})\) is a scalar:
$$A'^k_k = \frac{\partial x^i}{\partial y^k}\frac{\partial y^k}{\partial x^i}A^i_i = A^i_i,$$
(3)
Let \(\vec{v} = v^i\partial_i\) be a vector on \(M^n\).
(i) We define a new object by \(v_j = g_{ji}v^i\). By the transformation properties,
$$v'_k = \frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^k}g_{ji}\frac{\partial y^k}{\partial x_i}v^i = \frac{\partial x^j}{\partial y^k}g_{ji}v^i = \frac{\partial x^j}{\partial y^k}v_j,$$
thus, \(v_j\) defines a one-form.