Part 1
For the first inequation,
$$\tag{1}\label{1}
2(|a|^p + |b|^p) \leq |a + b|^p + |a - b|^p \leq 2^{p-1}(|a|^p + |b|^p), \quad p\ge2,
$$
the second inequality can be derived from the first one by the replacements $a\to a+b$ and $b\to a-b$. Thus, we concentrate on proving the first inequality, that is,
$$\tag{2}\label{2}
2(|a|^p + |b|^p) \leq |a + b|^p + |a - b|^p.
$$
We can obviously assume that $a\ne0$; if not, the inequality is trivially true. Without loss of generality, we also assume $|b|\le|a|$. Divide both side of \eqref{2} by $|a|$, and let $x:=b/a$ (for the case $|b|\ge|a|$, just divide by $|b|$ and let $x:=a/b$), then we only need to prove
$$
2(1+|x|^p) \le |1+x|^p+|1-x|^p \quad \text{for } |x|\le1.
$$
Now it is reduced to the following problem by the symmetry,
$$\tag{3}\label{3}
2(1+x^p) \le (1+x)^p+(1-x)^p \quad \text{for } 0\le x\le1.
$$
See here for the proof of \eqref{3}.
Part 2
For the second inequation,
$$\tag{4}\label{4}
|a + b|^p + |a - b|^p \leq 2^{p-1}(|a|^q + |b|^q)^{\frac{p}{q}}, \quad p \geq 2, \frac{1}{p} + \frac{1}{q} = 1,
$$
we only need to prove the following by the benefit of the second inequality in the first inequation \eqref{1},
$$
(|a|^q + |b|^q)^{\frac{1}{q}} \le (|a|^p + |b|^p)^{\frac{1}{p}} \quad \text{for } q\le p,
$$
which is an immediate result of the decrease of $p$-norm in finite dimensions, for whose proof one may refer to here or here.
For the third inequation,
$$
\big[2(|a|^p + |b|^p)\big]^{\frac{q}{p}} \leq |a + b|^q + |a - b|^q,
$$
one may just make replacements $a\to (a+b)/2$ and $b\to (a-b)/2$ in the second inequation \eqref{4} and compute
$$
|a|^p + |b|^p \le 2^{p-1} \bigg( \bigg| \frac{a+b}{2} \bigg|^q + \bigg| \frac{a-b}{2} \bigg|^q \bigg)^{\frac{p}{q}}
= 2^{-1} ( | a+b |^q + | a-b |^q )^{\frac{p}{q}},
$$
as required.
