1): We have: If $\angle{A} <90^o$ then: $\frac{BC}{\min{AD,BD,CD}} \geq \frac{BC}{R}$, where $R$ is circumcentre. $R=\frac{BC}{2sinA}$. So: $\frac{BC}{\min{AD,BD,CD}} \geq \frac{BC}{R}$ $\geq 2sinA$. If $\angle {A} \geq 90^o$ then $BC > AC,AB$, and: if $D$ is midpt of $BC$, then $minAD,BD,CD \leq \frac{BC}{2}$ and so $\frac{BC}{\min{AD,BD,CD}} \geq 2$. Let $M$ be midpoint of $BC$. if $D$ is inside $\triangle AMB$, $minAD,BD \le BM$, and if $D$ is inside $\triangle AMC$, $minAD,CD \le CM$. Therfore: $minAD,BD,CD \le CM= \frac{BC}{2}$.So: $\frac{BC}{\min{AD,BD,CD}} \geq 2$ QED

(2) It's obvious that there's no three points collinear (otherwise we get $k\geq 2>2\sin 70^{\circ}$). Also, we've the following Lemma: For any triangle $ABC$, $\frac{BC}{\min \{ BA,AC\} }\geq 2\sin \frac{A}{2}$. Suppose that $k\leq 2\sin 70^{\circ}$. Let $M$ is the smallest distance of two points among $A,B,C,D,E$. Let $AC\cap BD=T$. WLOG that $E$ is in the interior of $\triangle{ABT}$. By Lemma, we get $\angle{AEB},\angle{BED}\leq 140^{\circ}\implies \angle{AED}\geq 80^{\circ}$. By (1), we get that $\angle{ABC},\angle{BAD}\leq 70^{\circ}$. WLOG $\angle{BCD}\geq 110^{\circ}$. In triangle $BCE$, we've $\frac{BC}{\min \{ BE,EC\} }\geq 2\sin 40^{\circ}\implies BC\geq 2M\sin 40^{\circ}$. By law of cosines, $BD^2=BC^2+CD^2-2BC\times CD\times \cos (\angle{BCD})\geq BC^2+CD^2+2BC\times CD\times \sin 70^{\circ}\geq M^2(4\sin^2 40^{\circ} +1+4\sin 40^{\circ} \sin 70^{\circ})$. Hence, $BD^2\geq 4M^2\sin^2 70^{\circ}\implies BD\geq 2M\sin 70^{\circ}\implies 2\sin 70^{\circ} \geq k\geq 2\sin 70^{\circ}$. We get that if $k\leq 2\sin 70^{\circ}$, then $k=2\sin 70^{\circ}$. This gives $k\geq 2\sin 70^{\circ}$. For the equality case, we get that all equalities in the above bounds holds. This's not hard to verify that it happens when $ABCD$ is an isosceles trapezoid with $\angle{ABC}=\angle{BAD}=70^{\circ}$ and $E$ is the point in quadrilateral $ABCD$ that $DEC$ is an equilateral triangle.

Different solution for $(2)$ $(2)$ First of all, if $E$ lies on $AC$ or $BD$, then in fact we can get a bound of $2 > 2 \sin 70.$ Hence, assume from now on that $E$ is in the strict interior of $\triangle BCD$ and $\triangle BCA$, WLOG. We can also observe that if any triangle consisting of three of $A, B, C, D, E$ has an obtuse angle which is greater than $140$, we are done by Law of Sines within that triangle. By part $(1)$, if any of $\angle BAC, \angle ABC, \angle ACB, \angle BDC, \angle DCB, \angle DBC$ are greater than or equal to $70$ degrees, we are done. Hence, suppose that they all have measures in the range $(40, 70].$ Let $\theta_1 = \angle ABC, \theta_2 = \angle DCB.$ Observe that if $\angle BAD \le 70$ then $\angle ADC \ge 150$ and we're done. Analogously, we're also done if $\angle ADC \le 70.$ Therefore, WLOG assume that $\angle BAD, \angle CDA >70.$ Now, observe that either $\angle BAD \le 180 - \theta_1$ or $\angle ADC \le 180 - \theta_2.$ By symmetry, we will consider just the former. We then have that $\angle BAD \in [70, 180 - \theta_1]$ and so as $\theta_1 \le 70$, we've $\sin \angle BAD \ge \sin \theta_1.$ Therefore, $\frac{BD}{AD} \ge \frac{\sin \theta_1}{\sin (\theta_1 - 40)},$ where we also used $\angle ABD = \angle ABC - \angle DBC \le \theta_1 - 40.$ It's easily checked that for $\theta_1 \in (40, 70],$ $\frac{\sin \theta_1}{\sin (\theta_1 - 40)} \ge \frac{\sin 70}{\sin 30} = 2 \sin 70,$ and so we're done. $\square$