The answer to part (b) is that the normal closure of the word given in part (b) is always equal to $G$.

To see this, consider the quotient of $G$ by adding in this case a relation $a^{k}ta^{l}ta^{m}t=1$.

We will show that $a=1$, and then as $t^{4}=1$ it will also follow that $t=1$.

Let $b=tat^{-1},c=tbt^{-1},d=tat^{-1}$ with $bab^{-1}=a^{2},cbc^{-1}=b^{2},dcb^{-1}=c^{2},ada^{-1}=d^{2}$.

Write $a^{k}ta^{l}t^{-1}t^{2}a^{m}t^{-2}t^{3}=1$ so that $t=a^{k}b^{l}c^{m}=b^{k}c^{l}d^{m}=c^{k}d^{l}a^{m}=d^{k}a^{l}b^{m}$.

Case (1): Assume one of $k,l,m$ $\geqq0$ ; wlog assume $m\geqq0$.

Since $b^{m}ab^{-m}=a^{2^{m}}$ we have $b=tat^{-1}=$$d^{k}a^{l}b^{m}ab^{-m}a^{-l}d^{-k}=d^{k}a^{2^{m}}d^{-k}$. But $a^{2^{m}}da^{-2^{m}}=d^{2^{2^{m}}}$ so $b=d^{k}d^{-k2^{2^{m}}}a^{2^{m}}=d^{u}a^{2^{m}}$ where $u=k(1-2^{2^{m}})$.

Then $a^{2}=bab^{-1}=d^{u}a^{2^{m}}aa^{-2^{m}}d^{-u}=d^{u}ad^{-u}$. Using $ada^{-1}=d^{2}$ we have $a^{2}=d^{u}d^{-2u}a=d^{-u}a$ so that $a=d^{-u}$ and in particular $da=ad$. Then $ada^{-1}=d^{2}$ gives $d=1$.

Since $a$ and $d$ are conjugate $a=1$.

Case (2): Suppose $k,l,m$ all negative and not $0$.

Then $b=tat^{-1}=c^{k}d^{l}a^{m}aa^{-m}d^{-l}c^{-k}=c^{k}d^{l}ad^{-l}c^{-k}=c^{k}d^{l}d^{-2l}ac^{-k}=c^{k}d^{-l}ac^{-k}$ using $ad^{-l}a^{-1}=d^{-2l}$. Thus $c^{-k}bc^{k}=d^{-l}a$.

Since $-k$ is positive $c^{-k}bc^{k}=b^{2^{-k}}=b^{w}$ with $w=2^{-k}$. So $b^{w}=d^{-l}a$.

However, $b^{w}ab^{-w}=a^{2^{w}}$ and we see $d^{-l}ad^{l}=a^{2^{w}}$. Using $ad^{l}a^{-1}=d^{2l}$ we have $d^{-l}d^{2l}a=a^{2^{w}}$. Thus $d^{l}=a^{v}$ where $v=2^{w}-1$.

Consequently, from $a^{v}da^{-v}=d^{2^{v}}$ we can now deduce $d=d^{2^{v}}$ or $d^{2^{v}-1}=1$.

In any case $d$ has finite order, $n$. Since $d$ and $a$ are conjugate they have exactly the same order $n$.

It is now a standard argument originating with G. Higman that since $ada^{-1}=d^{2}$ we also have $d^{2^{n}-1}=1$ and so $n$ divides $2^{n}-1$. Then a simple number theory argument shows $n=1$.

Thus $a=1$.

The answer to part (b) is that the normal closure of the word given in part (b) is always equal to $G$.

To see this, consider the quotient of $G$ by adding in this case a relation $a^{k}ta^{l}ta^{m}t=1$.

We will show that $a=1$, and then as $t^{4}=1$ it will also follow that $t=1$.

Let $b=tat^{-1},c=tbt^{-1},d=tat^{-1}$ with $bab^{-1}=a^{2},cbc^{-1}=b^{2},dcd^{-1}=c^{2},ada^{-1}=d^{2}$.

Write $a^{k}ta^{l}t^{-1}t^{2}a^{m}t^{-2}t^{3}=1$ so that $t=a^{k}b^{l}c^{m}=b^{k}c^{l}d^{m}=c^{k}d^{l}a^{m}=d^{k}a^{l}b^{m}$.

Case (1): Assume one of $k,l,m$ $\geqq0$ ; wlog assume $m\geqq0$.

Since $b^{m}ab^{-m}=a^{2^{m}}$ we have $b=tat^{-1}=$$d^{k}a^{l}b^{m}ab^{-m}a^{-l}d^{-k}=d^{k}a^{2^{m}}d^{-k}$. But $a^{2^{m}}da^{-2^{m}}=d^{2^{2^{m}}}$ so $b=d^{k}d^{-k2^{2^{m}}}a^{2^{m}}=d^{u}a^{2^{m}}$ where $u=k(1-2^{2^{m}})$.

Then $a^{2}=bab^{-1}=d^{u}a^{2^{m}}aa^{-2^{m}}d^{-u}=d^{u}ad^{-u}$. Using $ada^{-1}=d^{2}$ we have $a^{2}=d^{u}d^{-2u}a=d^{-u}a$ so that $a=d^{-u}$ and in particular $da=ad$. Then $ada^{-1}=d^{2}$ gives $d=1$.

Since $a$ and $d$ are conjugate $a=1$.

Case (2): Suppose $k,l,m$ all negative and not $0$.

Then $b=tat^{-1}=c^{k}d^{l}a^{m}aa^{-m}d^{-l}c^{-k}=c^{k}d^{l}ad^{-l}c^{-k}=c^{k}d^{l}d^{-2l}ac^{-k}=c^{k}d^{-l}ac^{-k}$ using $ad^{-l}a^{-1}=d^{-2l}$. Thus $c^{-k}bc^{k}=d^{-l}a$.

Since $-k$ is positive $c^{-k}bc^{k}=b^{2^{-k}}=b^{w}$ with $w=2^{-k}$. So $b^{w}=d^{-l}a$.

However, $b^{w}ab^{-w}=a^{2^{w}}$ and we see $d^{-l}ad^{l}=a^{2^{w}}$. Using $ad^{l}a^{-1}=d^{2l}$ we have $d^{-l}d^{2l}a=a^{2^{w}}$. Thus $d^{l}=a^{v}$ where $v=2^{w}-1$.

Consequently, from $a^{v}da^{-v}=d^{2^{v}}$ we can now deduce $d=d^{2^{v}}$ or $d^{2^{v}-1}=1$.

In any case $d$ has finite order, $n$. Since $d$ and $a$ are conjugate they have exactly the same order $n$.

It is now a standard argument originating with G. Higman that since $ada^{-1}=d^{2}$ we also have $d^{2^{n}-1}=1$ and so $n$ divides $2^{n}-1$. Then a simple number theory argument shows $n=1$.

Thus $a=1$.