Can we say that $\left(a+b\right)^{\alpha}>a^{\alpha}+b^{\alpha}$ for all $a,b>0$ and $\alpha>1$?

For $\alpha\in\mathbb{N}$ we can use the Binomial and get: $$\left(a+b\right)^{n}=\sum_{k=0}^{n}{n \choose k}a^{k}b^{n-k}=\sum_{k=1}^{n-1}{n \choose k}a^{k}b^{n-k}+a^{n}+b^{n}>a^{n}+b^{n} $$

But what about rational and irrational powers?

Respectively, can we also say that $\left(a+b\right)^{\alpha}<a^{\alpha}+b^{\alpha}$ for all $\alpha<1$?

Sure. $$(a+b)^{\alpha} = a^{\alpha} + \int_0^b \alpha(a+x)^{\alpha-1}\,dx > a^{\alpha} + \int_0^b \alpha x^{\alpha-1}\,dx = a^{\alpha} + b^{\alpha}.$$

Dividing both sides by $(a+b)^{\alpha} $ it can be seen that the inequality is reduced the case when $a+b=1$. Therefore $0<a,b<1$ and $a^{\alpha} <a, b^{\alpha} <b$. And finally we have $$a^{\alpha} +b^{\alpha} <a+b=1=1^{\alpha}=(a+b)^{\alpha}$$ as desired. It can be proved in similar manner that the inequality is reversed if $\alpha<1$ and obviously there is equality if $\alpha=1$.

For the first one, you can write

$$(a+b)^\alpha = (a+b)(a+b)^{\alpha-1} = a(a+b)^{\alpha-1} + b(a+b)^{\alpha-1}$$

Now since $\alpha > 1$, $f(x) = x^{\alpha -1}$ is clearly an increasing function for $x >0$

And then since $a,b > 0$

$$f(a+b) > f(a) \implies (a+b)^{\alpha-1} > a^{\alpha-1}$$ $$f(a+b) > f(b) \implies (a+b)^{\alpha-1} > b^{\alpha-1}$$ And substituting into the above you have your result.

For your second question, when $\alpha < 1$, we can do exactly the same thing by noting that $f(x) = x^{\alpha-1}$ will now be a decreasing function for $x > 0$.