An exact base functor has the trivial universal delta functor

Fix a short exact sequence $0\to A\to B\to C\to 0$. Because $F$ is exact, [L1] gives an exact sequence $0\to F(A)\to F(B)\to F(C)\to 0.$ Therefore the homological family with degree zero $F$, higher degrees $0$, and zero connecting maps has the required long exact sequences and naturality squares, so it is a homological delta functor.

Let $S=(S_n,{\mathrm{\partial }}^S)$ be any homological delta functor and let $u_0:S_0\Rightarrow F$ be a natural transformation. Define $u_n=0$ for $n>0$. In every connecting square with $n>1$ this is automatic. For $n=1$, exactness of $S$ gives $\mathrm{im}({\mathrm{\partial }}_1^S)=\ker (S_0(A)\to S_0(B))$, while step 1.1 gives $\ker (F(A)\to F(B))=0$; naturality of $u_0$ therefore implies $u_0(A){\mathrm{\partial }}_1^S=0$, so the degree-one connecting square also commutes. The extension is unique because there is only one morphism into the zero object in each positive degree. Hence the trivial homological delta functor is universal by [L2].

The cohomological case is dual. Step 1.1 already gives exact sequences $0\to F(A)\to F(B)\to F(C)\to 0,$ so the family with $T^0=F$, $T^n=0$ for $n>0$, and zero connecting maps is a cohomological delta functor. Given any cohomological delta functor $S=(S^n,{\mathrm{\partial }}_S)$ and any $u^0:F\Rightarrow S^0$, define $u^n=0$ for $n>0$. Exactness of $S$ gives $\ker ({\mathrm{\partial }}_S^0)=\mathrm{im}(S^0(B)\to S^0(C))$, and exactness of $F$ makes $F(B)\to F(C)$ epic, so naturality of $u^0$ implies ${\mathrm{\partial }}_S^0{u}^0(C)=0$. Thus the connecting squares commute, and uniqueness is again immediate in positive degrees. Therefore the trivial cohomological delta functor is universal by [L2].