Respect to the second question, there is not analogous lemma for dim =13; but there is a lemma for dim = 15, as follows:
Let $Y^{15}$ be an orientable fifteen-manifold. Then we have
$w_{15}(Y^{15}) =w_{14}(Y^{15})=w_{13}(Y^{15}) = 0 $.
Proof. From the properties of the Wu classes we obtain for $Y^{15}$ that
$$\left\{ v_{{8}}=0,v_{{9}}=0,v_{{10}}=0,v_{{11}}=0,v_{{12}}=0,v_{{13}}=0,v_{{14}}=0,v_{{15}}=0 \right\}$$
Now from the Wu’s formula, we derive that
$$w_{{15}}=0$$
$$w_{{14}}={v_{{7}}}^{2}$$
$$w_{{13}}={\it Sq}^{{6}} \left( v_{{7}} \right) $$
From other side we know that
$$v_{{7}}={w_{{1}}}^{2}w_{{2}}w_{{3}}+w_{{1}}{w_{{3}}}^{2}+w_{{1}}w_{{2}
}w_{{4}}
$$
but given that $Y^{15}$ is orientable, it is to say $w_1 =0$; we obtain that $v_7=0$.
For hence we have that
$$w_{{15}}=0$$
$$w_{{14}}={v_{{7}}}^{2} = 0^2 = 0$$
$$w_{{13}}={\it Sq}^{{6}} \left( v_{{7}} \right)= {\it Sq}^{{6}} \left( 0\right)=0 $$
And then our lemma is proved.
Do you agree?