Note that
$$\dfrac{(-1)^{k-1}}k = \int_0^1 (-x)^{k-1}dx$$
and
$$\dfrac1n = \int_0^1 y^{n-1}dx$$

\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\
& = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\
& = \sum_{n=1}^{\infty} \int_0^1 \dfrac{(-x)^{n-1}}{1+x}dx \int_0^1 y^{n-1} dy\\
& = \int_0^1 \int_0^1\sum_{n=1}^{\infty} \dfrac{(-xy)^{n-1}}{1+x}dx dy\\
& = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dx dy\\
& = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dy dx\\
& = \int_0^1 \dfrac{\log(1+x)}{x(1+x)} dx\\
& = \int_0^1 \dfrac{\log(1+x)}{x} dx - \int_0^1 \dfrac{\log(1+x)}{(1+x)} dx\\
& = \dfrac{\zeta(2)}2 - \dfrac{\log^2 2}2
\end{align}

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For the second one,

$$A(1,2) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n $$
$$\dfrac{(-1)^{k-1}}{k^2} = \int_0^1 (-x)^{k-1} dx \int_0^1 z^{k-1} dz = (-1)^{k-1} \int_0^1 \int_0^1 (xz)^{k-1} dx dz$$
\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}dxdz \int_0^1 y^{n-1} dy\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+xz} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+xz)(1+xyz)} dx dy dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz(1+xz)} dx dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz - \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{1+xz} dx dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz- \int_0^1 \dfrac{\log^2(1+z)}{2z} dz\\
& = \dfrac34 \zeta(3) - \dfrac{\zeta(3)}8\\
& = \dfrac58 \zeta(3)
\end{align}

---

$$A(2,1) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2} $$
\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2} & = \int_0^1 \int_0^1 \int_0^1 \sum_{k=1}^{\infty} \sum_{n=1}^k (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+x} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+x)(1+xyz)} dx dy dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xy)}{(1+x)(xy)} dx dy\\
& = \zeta(3) - \dfrac{\zeta(2) \log 2}2
\end{align}

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In general, if I have not made any mistake, this can be extended to $A(p,q)$.
$$A(p,q) = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$