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No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ is $\mathbb Q^{alg}[x]$ and has a bijection to $\prod_{n=1}^\infty \mathbb Q^{alg}$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.

No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ is $\mathbb Q^{alg}[[x]]$ and has a bijection to $\prod_{n=1}^\infty \mathbb Q^{alg}$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.

No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ has a bijection to $\prod_{n=1}^\infty \mathbb Q$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.

No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ is $\mathbb Q^{alg}[x]$ and has a bijection to $\prod_{n=1}^\infty \mathbb Q^{alg}$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.

No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ has a bijection to $\prod_{n=1}^\infty \mathbb Q$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.

No, since $K(\hat{A})$ is usually transcendental over $K(A)$.

Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ has a bijection to $\prod_{n=1}^\infty \mathbb Q$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.