Write the ellipse in standard form: $\frac{x^2}{1/3}+\frac{y^2}{1/4}=1$, so $a^2=1/3$, $b^2=1/4$. Then $e=\sqrt{1-b^2/a^2}=\sqrt{1-3/4}=1/2$.

Apply the tangency condition $c^2=a^2{m}^2+b^2$ to $y=-\frac{\alpha }{4}x+\frac{\sqrt{7}}{4}$: $\frac{7}{16}=\frac{1}{3}\cdot \frac{{\alpha }^2}{16}+\frac{1}{4}$, giving ${\alpha }^2=9$. Taking $\alpha =3$ (so P is in the first quadrant): tangent $3x+4y=\sqrt{7}$.

Point of contact: comparing with $3x{x}_1+4y{y}_1=1$ gives $\frac{3x_1}{3}=\frac{4y_1}{4}=\frac{1}{\sqrt{7}}$, so $P=\left(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}}\right)$.

Focal distance using $r=a+e{x}_1$ (for the closer focus on the negative-x side, $r=a-e{x}_1$; for the farther, $r=a+e{x}_1$): $r_{+}=\frac{1}{\sqrt{3}}+\frac{1}{2}\cdot \frac{1}{\sqrt{7}}=\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.