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?