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Solution

For $r=a(b+a)^*$:

This represents all strings over $\{a,b\}$ that start with $a$.

So,

$L(r)=\{a,aa,ab,aaa,aab,aba,\ldots\}$

For $s=a(a+b)^+$:

Here, after first $a$, at least one more symbol from $\{a,b\}$ must be present.

So, $s$ represents all strings over $\{a,b\}$ that start with $a$ and have length at least $2$.

Therefore,

$L(s)\subseteq L(r)$

For $t=aa^*b$:

This represents strings of the form:

$ab,aab,aaab,aaaab,\ldots$

So, every string of $L(t)$ starts with $a$ and has length at least $2$.

Therefore,

$L(t)\subseteq L(s)$

Hence,

$L(r)\supseteq L(s)\supseteq L(t)$