Call the triangle $ABC$, and the altitude $AD$.
(1) construct a triangle with perimeter length 115 and with the altitude length 70.
That is impossible. $ADB$ is a right-angled triangle, so $AB>AD$. Similarly, $AC>AD$. So the perimeter must be greater than $2AD$.
(2) construct a triangle with $\angle B=45\circ$, $\angle C=60\circ$ and $AD=76$.
We have $AB\sin\angle B=AD$, which gives us $AB$. Similarly $AC\sin\angle C=AD$, which gives us $AC$. Similarly, $AB\cos\angle B+AC\cos\angle C=BC$. So we have all the side lengths and constructing the triangle is trivial.
(3) construct a triangle with perimeter 135, $AD=55$ and $AB=AC$. Suppose $BC=x$. Then $AB=AC=\sqrt{55^2+\left(\frac{x}{2}\right)^2}$. So we have perimeter $135=x+2\sqrt{55^2+\left(\frac{x}{2}\right)^2}$. So we can solve for $x$ (approx 22.7) and hence get the side lengths.