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I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

Here, $E:=B(X)$, $F:=X\otimes_{alg} X^*$ as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $T\mapsto\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$  ), and $V:=X\otimes_{alg} X^*$ as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $u\mapsto Tu:=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$).

Thanks to the above general fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.

Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.

I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that the $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

In your situation, $E:=B(X)$, $F:=X\otimes_{alg} X^*$, seen as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $T\mapsto\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$), and $V:=X\otimes_{alg} X^*$, seen as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $u\mapsto Tu:=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$).

Thanks to the above general fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.

Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.

I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

Here, $E:=B(X)$, $F:=X\otimes_{alg} X^*$ as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$ ), and $V:=X\otimes_{alg} X^*$ as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $Tu=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$).

Thanks to the above fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.

Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.

I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

Here, $E:=B(X)$, $F:=X\otimes_{alg} X^*$ as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $T\mapsto\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$ ), and $V:=X\otimes_{alg} X^*$ as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $u\mapsto Tu:=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$).

Thanks to the above general fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.

Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.

I think yes, for general reasons. If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

Here $E=B(X)$, $F=X\otimes_{alg} X^*$ as linear space of linear forms on $B(X)$, and $V:=X\otimes_{alg} X^*$ as the subset of $B(X)$ of finite rank operators.

Indeed, let $f$ be a non-zero element of $F$. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the element $T=\sum_{j=1}^m y_j\otimes y^*_j\in X\otimes_{alg} X^*$ (that is the finite rank operator $T:u\mapsto \sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$) is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.

I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).

Here, $E:=B(X)$, $F:=X\otimes_{alg} X^*$ as a linear space of linear forms on $B(X)$ (that is, $f=\sum_{i=1}^m x_i\otimes x^*_i$ corresponds to the linear form $\langle T,f\rangle:= \sum_{i=1}^m \langle Tx_i, x^*_i\rangle$ on $B(X)$ ), and $V:=X\otimes_{alg} X^*$ as the subset of $B(X)$ of finite rank operators (that is, $T=\sum_{j=1}^m y_j\otimes y^*_j$ corresponds to the finite rank operator $Tu=\sum_{j=1}^m y_j\langle u,y^*_j \rangle$ on $X$).

Thanks to the above fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.

Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum_{i=1}^m x_i\otimes x^*_i$ be a representation of $f$ with minimum $m$. This implies that $x_1,\dots,x_m$ are linearly independent elements of $X$ and $x_1^*,\dots,x_m^*$ are linearly independent elements of $X^*$. Then there are $y_1^*,\dots,y_m^*$ in $X^*$ such that $\langle x_i,y^*_j\rangle=\delta_{ij}$, and there are $y_1,\dots,y_m$ in $X$ such that $\langle y_i,x^*_j\rangle=\delta_{ij}$. Then the operator $T=\sum_{j=1}^m y_j\otimes y^*_j$ is such that $\langle T,f\rangle:=\sum_{i=1}^m \langle Tx_i, x^*_i\rangle=\sum_{ij}\langle y_j,x^*_i \rangle\langle x_i,y^*_j \rangle=\sum_{ij}\delta_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$.