For the first limit, I think you break the limit up like this:
$$\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt x}=\lim_{x \to 0} \sqrt x\frac{\ln(x+1)}{x}$$
We know that as $x \to 0$, we have $\sqrt x \to 0$ and from your given limits, $\frac{\ln(x+1)}{x} \to 1$, so we get $0\cdot 1=0$.
For the second limit, I think we should do $x=t^2$ to get $x^2$ on the bottom. That way, there is a square on both the numerator and denominator:
$$\lim_{t \to \infty} \frac{\ln^2(t^2)}{t^2}=\frac{(2\ln(t))^2}{t^2}=4\left(\frac{\ln(t)}{t}\right)^2$$
By your given limits, we have $\frac{\ln(t)}{t} \to 0$, so by squaring that and multiplying by $4$, we get the limit's value to be $0$.
Best Answer
OK, one. Here comes: $$ \frac{\sqrt{u}-\sqrt{v}}{\sqrt[3]{u}-\sqrt[3]{v}} = \frac{u-v}{\sqrt{u}+\sqrt{v}}\cdot\frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{u-v}= \frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{\sqrt{u}+\sqrt{v}}\underset{u,v\to1}{\longrightarrow}\frac{1+1+1}{1+1}. $$ Equivalently, when $x\to0$, $(1\pm x)^a=1\pm ax+o(x)$ hence $$ \frac{(1+x)^a-(1-x)^a}{(1+x)^b-(1-x)^b}=\frac{1+ax-(1-ax)+o(x)}{1+bx-(1-bx)+o(x)}=\frac{2ax+o(x)}{2bx+o(x)}\underset{x\to0}{\longrightarrow}\frac{a}b. $$