Suppose $A=\{x_1,\dots,x_k\}$. Then $H_0(A)=\mathbb{Z}^k$ and $H_i(A)=0$ for $i>0$.
Whether $X=S^2$ or $T^2$ we have $H_0(X)\cong\mathbb{Z}$, and like Matt N said in his comment in either case $H_0(X,A)\cong\tilde{H}_0(X/A)=0$.
If $X=S^2$ then $H_1(X)=0$ so the l.e.s. has a portion like $$0\rightarrow H_1(X,A)\rightarrow \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0 $$ and so $H_1(S^2,A)\cong\mathbb{Z}^{k-1}$.
If $X=T^2$ then $H_1(X)=\mathbb{Z}^2$, so we have $$ 0\rightarrow \mathbb{Z}^2\rightarrow H_1(X,A)\stackrel{\partial}{\rightarrow} \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0$$
Then $\ker\partial\cong\mathbb{Z}^2$ and its image is $\cong\mathbb{Z}^{k-1}$. I believe this is enough to conclude $H_1(T^2,A)\cong \mathbb{Z}^{k+1}$